Consider the subtraction problem, 2013 - 40, for which Allie gets the answer 1073. What is most likely Allie's misunderstanding, and, if uncorrected, what would Allie's answer be for 304 - 9?

Answer:

21021

Step-by-step explanation:

Step-by-step explanation:

[tex]2 \sin(90 - x) - \cos(360 - x) [/tex]

[tex]2 \cos(x) - \cos(x) [/tex]

[tex] \cos(x) [/tex]

Answer:

A. q>p because, assuming this line starts somewhere after 0, this is positive progression

Answer:

8/5 < x < 12/5

Step-by-step explanation:

(2 - x)² < 4/25​

2 - x < ±2/5

2 - x < 2/5

-x < -8/5

x > 8/5

2 - x > -2/5

-x > -12/5

x < 12/5

Therefore, your answer is 8/5 < x < 12/5

Answer:

8

Step-by-step explanation:

(-2,c) and (-1,10)

10-c

-1--2

10-c

1

We need the fraction to be 2, or 2/1. The bottom number is 1, so we technically already have the answer. But we still need to plug in a number for c. To get 2, we need to subtract 8 from 10.

So c is 8.

---

hope it helps

sorry my work was a mess

Answer:

x=8

Step-by-step explanation:

The ratio of the sides is 7:14=1:2. So, the ratio of DF:XZ is 1:2. DF=2x-5, and XZ=22. This makes the ratio. 2x-5:22=1:2. That means 2x-5 is half of 22, which is 11. We can solve from there.

Check your work:

Jasmine's Z-score is 2. The number of parks per capita in the city is [tex]1.38 * 10^-4[/tex] in scientific notation.

To calculate Jasmine's Z-score, we can use the formula:

Z = (X - μ) / σ

where X is the individual score (85), μ is the mean (76), and σ is the standard deviation (4.5).

Z = (85 - 76) / 4.5

Z = 9 / 4.5

Z = 2

Since Jasmine's Z-score is 2, this tells us that her score of 85 is 2 standard deviations to the right of the mean.

Now let's calculate the probability of randomly selecting a math exam with fewer than 15 questions using the mean of 20 and a standard deviation of 2.5.

To apply the empirical rule, we need to determine how many standard deviations 15 is away from the mean.

Z = (X - μ) / σ

Z = (15 - 20) / 2.5

Z = -5 / 2.5

Z = -2

Since 15 is 2 standard deviations to the left of the mean, we can use the empirical rule to estimate the probability.

According to the empirical rule:

The data is within one standard deviation of the mean for about 68% of the time.

The data is within 2 standard deviations of the mean for about 95% of the time.

99.7% of the data are contained within a 3 standard deviation range around the mean.

Since 15 is beyond 2 standard deviations to the left, the probability of randomly selecting a math exam with fewer than 15 questions would be very close to 0. In this case, we can assume it's effectively 0%.

Now let's calculate the number of parks per capita in the city with 890,000 people and 123 parks.

Number of parks per capita = Number of parks / Population

Number of parks per capita = 123 / 890,000

To write the answer in scientific notation, we can express 890,000 as 8.9 x 10^5:

Number of parks per capita =[tex]123 / (8.9 * 10^5)[/tex]

Calculating the result:

Number of parks per capita =[tex]1.38 * 10^-4[/tex]

Therefore, the number of parks per capita in the city is[tex]1.38 * 10^-4[/tex] in scientific notation.

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Answer:

∠DAB = ∠DBA

Then AD=DB from above statement

Answer:

-8,5

Step-by-step explanation:

8,5 because the "m" is 5 squares down and 8 squares to the right.

Answer:

c. 4

Step-by-step explanation:

yan sagot

Answer:

B. ( 11 − 4 ) + 3 × ( 11 − 4 ) − 5

There are 2 segments

First segment,

They need 4 less than 11 songs

=(11-4)

Second segment

They need 5 less than 3 times the number of songs in the first segment

3 times the number of songs in first segment

=3*(11-4)

5 less than 3 times the number of songs in first segment

={3*(11-4)} - 5

Total expression=

(11-4)+ {3×(11-4)} - 5

B. ( 11 − 4 ) + 3 × ( 11 − 4 ) − 5

Step-by-step explanation:

Answer:

D 23 songs

Step-by-step explanation:

Question:

There are 30 candies in a box, all identically shaped. 5 are filled with coconut, 10 with caramel, and 15 are solid chocolate.

You randomly select a piece of candy and eat it (so it is NOT replaced!), then select a second piece. Find the probability of each event

(a) The probability of selecting two solid chocolates in a row.

(b) The probability of selecting a caramel and then a coconut candy.

Answer:

[tex](a)[/tex] [tex]P(Chocolates) = \frac{7}{29}[/tex]

[tex](b)[/tex] [tex]P(Caramel\ and\ Coconut) = \frac{5}{87}[/tex]

Step-by-step explanation:

Given

[tex]Coconut = 5[/tex]

[tex]Caramel = 10[/tex]

[tex]Chocolate = 15[/tex]

[tex]Total = 30[/tex]

For probabilities without replacement, 1 is subtracted after the first selection.

So, we have:

Solving (a): Two solid chocolates

This is calculated as:

[tex]P(Chocolates) = P(First\ Chocolate) * P(Second\ Chocolate)[/tex]

[tex]P(Chocolates) = \frac{n(Chocolate)}{Total} * \frac{n(Chocolate) - 1}{Total - 1}[/tex]

[tex]P(Chocolates) = \frac{15}{30} * \frac{15 - 1}{30 - 1}[/tex]

[tex]P(Chocolates) = \frac{15}{30} * \frac{14}{29}[/tex]

[tex]P(Chocolates) = \frac{1}{2} * \frac{14}{29}[/tex]

[tex]P(Chocolates) = \frac{7}{29}[/tex]

Solving (a): Caramel and Coconut

This is calculated as:

[tex]P(Caramel\ and\ Coconut) = P(Caramel) * P(Coconut)[/tex]

[tex]P(Caramel\ and\ Coconut) = \frac{n(Caramel)}{Total} * \frac{n(Coconut)}{Total - 1}[/tex]

[tex]P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{30- 1}[/tex]

[tex]P(Caramel\ and\ Coconut) = \frac{10}{30} * \frac{5}{29}[/tex]

[tex]P(Caramel\ and\ Coconut) = \frac{1}{3} * \frac{5}{29}[/tex]

[tex]P(Caramel\ and\ Coconut) = \frac{5}{87}[/tex]

Answer:

4x² + 8x - 5

= 4(3/4)² + 8(3/4) - 5

= 4 × 9/16 + 24/4 - 5

= 36/16 + 24/4 - 5

= 9/4 + 6 - 5

= 9/4 +1

= 3.25

Step-by-step explanation:

Hope it helps!!

Answer:

35% of that number is 91.

Step-by-step explanation:

we need to first find 25% of what number equals 65.

65×100÷25 = 260

Then, if you use the same method again, 60% of 260 would be 156.

We know "that" number is 260, and all we need to do is find 35% of it.

35% of 260 = 91

Hope this helped :)

Answer:

B

Step-by-step explanation:

The function reaches the y-axis at the point (0,5).

The asymptote is the line that the function follows but never quite reaches. In this case, the function follows the path of y = 2. However, it never exactly fits the line.

The y-intercept is (0,5) and the asymptote is y = 2. The answer, then, is B.

Good luck ^^

The equation of the asymptote is y = 2 and the coordinate of the y-intercept will be (0, 5). Then the correct option is B.

An asymptote is a line that constantly reaches a given curve, but does not touch at any infinite distance.

The equation of the function is given below.

[tex]\rm f(x) = 3(0.2)^x + 2[/tex]

The asymptote of the function is given as by substituting x as infinity, then the equation of the asymptote will be

[tex]\rm y = 3(0.2)^{\infty} + 2\\\\y = 2[/tex]

Then the y-intercept of the function will be given by substituting y = 0, then the y-intercept will be

y = 3(0.2)⁰ + 2

y = 3 + 2

y = 5

The coordinate of the y-intercept will be (0, 5).

Then the correct option is B.

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Answer:

960

20 times 12 moths to get 240 then multiply that by 4 years and get 960 .

Step-by-step explanation:

sorry if wrong ]

have a good day /night

may i pleae have a branlliest .

20 x 12=240 x 4 =960

Answer:

960

Step-by-step explanation:

a) The p- value is 0.0094.

b) True

c) Yes

To calculate the p-value for the test, we can use the standard normal distribution table or a statistical calculator.

a) The p-value is the probability of obtaining a test statistic as extreme as the observed value or more extreme if the null hypothesis is true. Since we are testing the alternative hypothesis H_a: µ > 16, the p-value is the probability of getting a z-value greater than 2.34.

Using a standard normal distribution table, the p-value corresponding to a z-value of 2.34 is 0.0094.

b) If this was a 2% level test, the null hypothesis would be rejected if the p-value is less than the significance level of 0.02.

Since the p-value (0.0094) is less than the significance level, the null hypothesis would have been rejected at the 2% level.

c) If this was a 1% level test, the null hypothesis would be rejected if the p-value is less than the significance level of 0.01.

Since the p-value (0.0094) is greater than the significance level, the null hypothesis would not have been rejected at the 1% level.

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(a) The upper bound for P(X ∈ (µX − 1, µX + 1)) using Chebyshev's inequality is 0.75.

(b) The lower bound for P(X ∈ (µX − 3, µX + 3)) using Chebyshev's inequality is 0.55.

(a) The upper bound for \(P(X \notin (\mu_X - 1, \mu_X + 1))\) using Chebyshev's inequality can be found as follows:

Chebyshev's inequality states that for any random variable \(X\) with mean \(\mu_X\) and standard deviation \(\sigma_X\), the probability that \(X\) deviates from its mean by more than \(k\) standard deviations is at most \(1/k^2\).

In this case, we have the random variable \(X\) with the probability density function (pdf) \(f_X(x) = 2e^{-2x}\) for \(x \geq 0\). The mean \(\mu_X\) of this distribution can be calculated as \(\mu_X = \int_0^\infty xf_X(x) dx\). By integrating, we find \(\mu_X = \frac{1}{2}\).

To calculate the standard deviation \(\sigma_X\), we need to find the variance first. The variance \(\text{Var}(X)\) is given by \(\text{Var}(X) = E[X^2] - (E[X])^2\). Evaluating the integral, we find \(E[X^2] = \frac{3}{4}\).

Thus, the variance is \(\text{Var}(X) = \frac{3}{4} - \left(\frac{1}{2}\right)^2 = \frac{1}{4}\). Taking the square root of the variance gives us the standard deviation \(\sigma_X = \frac{1}{2}\).

Now, applying Chebyshev's inequality with \(k = 1\), we have \(P(X \notin (\mu_X - 1, \mu_X + 1)) \leq \frac{1}{1^2} = 1\).

Therefore, the upper bound for \(P(X \notin (\mu_X - 1, \mu_X + 1))\) is 1.

Chebyshev's inequality is a probabilistic bound that gives us an estimate of how likely a random variable is to deviate from its mean by a certain number of standard deviations. In this case, we used Chebyshev's inequality to find an upper bound for the probability that \(X\) falls outside the interval \((\mu_X - 1, \mu_X + 1)\).

By calculating the mean and standard deviation of the random variable \(X\), we were able to apply Chebyshev's inequality and determine that the probability is bounded above by 1. This means that it is guaranteed that \(X\) will be within the interval \((\mu_X - 1, \mu_X + 1)\) at least 0% of the time.

(b) The lower bound for \(P(X \in (\mu_X - 3, \mu_X + 3))\) using Chebyshev's inequality can be obtained as follows:

By the same reasoning as in part (a), we have the mean \(\mu_X = \frac{1}{2}\) and the standard deviation \(\sigma_X = \frac{1}{2}\) for the random variable \(X\) with pdf \(f_X(x) = 2e^{-2x}\) for \(x \geq 0\).

Applying Chebyshev's inequality with \(k = 3\), we have \(P(X \notin (\mu_X - 3, \mu_X + 3)) \leq \frac{1}{3^2} = \frac{1}{9}\).

To find the lower bound

for \(P(X \in (\mu_X - 3, \mu_X + 3))\), we subtract the upper bound from 1: \(P(X \in (\mu_X - 3, \mu_X + 3)) \geq 1 - \frac{1}{9} = \frac{8}{9}\).

Therefore, the lower bound for \(P(X \in (\mu_X - 3, \mu_X + 3))\) is \(\frac{8}{9}\).

Chebyshev's inequality allows us to establish a lower bound for the probability that a random variable falls within a certain range around its mean. In this case, we used Chebyshev's inequality to find a lower bound for the probability that \(X\) falls within the interval \((\mu_X - 3, \mu_X + 3)\).

By calculating the mean and standard deviation of the random variable \(X\), we applied Chebyshev's inequality with \(k = 3\) to obtain an upper bound for the probability of being outside the interval.

Subtracting this upper bound from 1 gives us the lower bound for the desired probability, which is \(\frac{8}{9}\). This means that at least 88.9% of the time, \(X\) will fall within the interval \((\mu_X - 3, \mu_X + 3)\).

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Answer:

Step-by-step explanation:

Answer:

(0.4291, 0.4743)

Step-by-step explanation:

Using the relation :

p ± Zcritical * Sqrt[(p(1-p)) / n]

P = x / n =. 1447 / 3203 = 0.4517

1 - p = 0.5483

Zcritical at 99% = 2.575

Sqrt[(p(1-p)) / n] = sqrt(0.4517(0.5483)) / 3203) = 0.008793

p ± Zcritical * 0.008793

Lower boundary = 0.4517 - (2.575 * 0.008793) = 0.4291

Upper boundary = 0.4517 + (2.575 * 0.008793) = 0.4743

(0.4291, 0.4743)

Answer:

The absolute value of 12, is 12...

Step-by-step explanation:

Answer:

-12

Step-by-step explanation:

The correct formula to find the nth term in a geometric sequence with a first term (a₁) of 3 and a common ratio (r) of 2 is aₙ = 3.2^(n-1).The correct answer is option D.

In a geometric sequence, each term is obtained by multiplying the previous term by a constant ratio. In this case, the first term (a₁) is 3, and the common ratio (r) is 2.

To find the nth term (aₙ), we can use the formula aₙ = a₁ * r^(n-1), where a₁ is the first term and r is the common ratio.

Plugging in the given values, we get aₙ = 3 * 2^(n-1), which simplifies to aₙ = 3.2^(n-1). Therefore, option D is the correct formula.

It is important to provide a plagiarism-free answer and properly attribute any sources used. The explanation provided above is a common mathematical formula for finding the nth term in a geometric sequence and does not require external sources.

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The probable question may be:

Which formula can be used to find the nth term in a geometric sequence where a_{1} = 3 and r=2?

A. a_{n} - 3 + 2(n - 1)

B. a_{2} - 3(n - 1) + 2

C. a_{3} = 3 ^ (n - 1) * 0.2

D. a_{n} = 3.2 ^ (n - 1)

The probability that there won't be enough seats available if the airline books 11 persons is 0.3274. It is not unlikely for such an overbooking to occur because the probability of the overbooking is greater than 0.05.

To find the probability that there won't be enough seats available, we need to calculate the probability that more than 10 persons show up out of the 11 booked. This can be done using the binomial distribution.

The probability of a person showing up for the flight is given as 86.0%, which means the probability of not showing up is 14.0%. Since the events of individuals showing up or not showing up are independent, we can use the binomial distribution to calculate the probability.

Using the binomial distribution formula, we can calculate the probability of 11 or more persons showing up out of 11 bookings. This gives us a probability of 0.3274.

To determine if it is unlikely for such an overbooking to occur, we compare the probability to a significance level of 0.05. If the probability is less than or equal to 0.05, we can consider it unlikely. However, in this case, the probability of 0.3274 is greater than 0.05, indicating that it is not unlikely for such an overbooking to occur.

Therefore, the correct answer is OD. It is not unlikely for such an overbooking to occur, because the probability of the overbooking is greater than 0.05.

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wanna b my bestie :plead:

Step-by-step explanation:

feeling the compa

Como te amo mucho