Helpppppppp pleaseeeeee I will give brainliest if you knowww pleaseeee helpppppppp

Answer:

One

Step-by-step explanation:

3x-5=-3 add 5 to each side.

3x-5+5=-3+5   simplify.

3x=2       divide each side by three.

3x/3=2/3     simplify

x=2/3

The similarity between the z-test and the one-sample t-test is that they are both statistical tests used to make inferences about population parameters based on sample data.

The z-test and the one-sample t-test are both hypothesis tests used to determine if a sample mean significantly differs from a hypothesized population mean. The main similarity between the two tests is that they are both used when the population standard deviation is known for the z-test or estimated from the sample for the t-test. Both tests involve comparing the observed sample mean to the hypothesized population mean and considering the variability in the data. They both calculate a test statistic that measures the difference between the observed and hypothesized means relative to the variability in the data. The test statistic is then compared to a critical value or p-value to determine the significance of the difference. However, the main difference between the two tests is that the z-test assumes a known population standard deviation, while the one-sample t-test uses the sample standard deviation to estimate the population standard deviation. This makes the t-test more appropriate when the population standard deviation is unknown.

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Answer:

10sqrt{2}

Step-by-step explanation:

This is a 45-45-90 right triangle. This means that the two legs are both the same length, call it x, and the hypotenuse is square root 2 times the given leg.  There's a proof for this, but it's long, and you can find it online.    

Answer:

7. 8 I think (typing more for the limit)

Answer:

7.8

Step-by-step explanation:

The probability that a randomly selected man weighs less than 163 pounds is approximately 0.7422 or 74.22%.

We are given that the weights of adult males are normally distributed with a mean of 150 pounds and a standard deviation of 20 pounds.

(a) To find the probability that a randomly selected man will weigh less than 163 pounds, we need to calculate the area under the normal curve to the left of 163 pounds.

To do this, we can standardize the value using the formula for z-score:

z = (x - μ) / σ

where x is the value we want to standardize, μ is the mean, and σ is the standard deviation.

In this case, x = 163 pounds, μ = 150 pounds, and σ = 20 pounds.

Calculating the z-score:

z = (163 - 150) / 20

z = 13 / 20

z = 0.65

Now, we can use a standard normal distribution table or a calculator to find the corresponding probability for a z-score of 0.65. The probability of a man weighing less than 163 pounds is the area to the left of the z-score of 0.65.

Looking up the z-score in the standard normal distribution table, we find that the corresponding probability is approximately 0.7422.

Therefore, the probability that a randomly selected man will weigh less than 163 pounds is approximately 0.7422, or 74.22%.

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Answer:

20

Step-by-step explanation:

4+4+6+6=20

(JUST WHAT SHE SAID)(SORRY FOR THE CAPS HERE)

Answer:

Thank you so much

Step-by-step explanation:

v₁ = [[-1], [-2], [-2]] and v₂ = [[-4], [8], [-6]] are the vectors that satisfy the given conditions.

To write vector y as the sum of a vector v₁ in W and a vector v₂ orthogonal to W, we need to find the orthogonal projection of y onto the subspace W spanned by u₁ and u₂.

y = [[-5], [6], [-8]]

u₁ = [[1], [2], [2]]

u₂ = [[6], [2], [-5]]

To find v₁, we'll use the formula for the orthogonal projection

v₁ = ((y · u₁) / (u₁ · u₁)) × u₁

where "·" represents the dot product.

Calculating the dot products

y · u₁ = (-5 × 1) + (6 × 2) + (-8 × 2) = -5 + 12 - 16 = -9

u₁ · u₁ = (1 × 1) + (2 × 2) + (2 × 2) = 1 + 4 + 4 = 9

Substituting the values

v₁ = ((-9) / 9) × [[1], [2], [2]] = [[-1], [-2], [-2]]

Now, to find v₂, we'll subtract v₁ from y

v₂ = y - v₁ = [[-5], [6], [-8]] - [[-1], [-2], [-2]] = [[-4], [8], [-6]]

Therefore, we can write y as the sum of v₁ and v₂

y = v₁ + v₂ = [[-1], [-2], [-2]] + [[-4], [8], [-6]] = [[-5], [6], [-8]]

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Answer: 95%

Step-by-step explanation: 0.9 = 90%

Answer:

i think its 7/10

Step-by-step explanation:

Since there are a total of 15 shirts,

5 red shirts+4 white shirts= 9 shirts total

3 blue shirts+3 purple shirts= 6 shirts total

The odds that you would pick a red shirt and then a white are 3:2

3 being the 9 red and white shirts combined.

2 being the 6 purple and blue shirts combined.

9:6 simplified=3:2

9/3=3

6/3=2

3:2

Answer: The odds that you would pick a red shirt then a white shirt would be 3:2 or 60%

Answer:

Answer is D

Step-by-step explanation:

hope that helps

Answer:

7/8 if u want more if u want less then it 4/9

Step-by-step explanation:

Answer:

Brian will have $12 dollars left.

Step-by-step explanation:

30 - 18 = 12

Answer:

c

Step-by-step explanation:

Answer:

k = 3.

Step-by-step explanation:

If they are collinear the slope of AB = the slope of BC, so :-

(0- (-2)) / (2 - 0) = (k - 0) / (5 - 2)

2/2 = k/3

1 = k/3

k = 3.

According to the question the following scenarios are as follows :

(a) To find the probability that A and B are next to each other in a line of 20 people, we can treat A and B as a single entity. This reduces the problem to arranging 19 entities (A and B together with the other 18 people) in a line. The total number of arrangements of 19 entities in a line is [tex]$19!$[/tex] .

Now, within the 19 entities, A and B can be arranged in [tex]$2! = 2$[/tex] ways (A followed by B or B followed by A). For each arrangement of A and B, the remaining 18 people can be arranged in [tex]$18!$[/tex] ways.

Therefore, the total number of arrangements where A and B are next to each other is [tex]$2 \cdot 18!$[/tex] .

The total number of possible arrangements of 20 people in a line is [tex]$20!$[/tex] .

The probability of A and B being next to each other is given by:

[tex]\[P(A \text{ and } B \text{ are next to each other}) = \frac{2 \cdot 18!}{20!}\][/tex]

(b) If the people are randomly arranged in a circle, we can fix one person (let's say A) at a specific position. Then, we arrange the remaining 19 people in a line, which can be done in [tex]$19!$[/tex] ways.

However, since the circle allows for rotations, each arrangement can be rotated in 20 different ways (corresponding to the different positions of A as the fixed person).

Therefore, the total number of arrangements where A and B are next to each other in a circle is [tex]$20 \cdot 19!$[/tex] .

The total number of possible arrangements of 20 people in a circle is [tex]$(20 - 1)!$[/tex] , since one person is fixed.

The probability of A and B being next to each other in a circle is given by:

[tex]\[P(A \text{ and } B \text{ are next to each other in a circle}) = \frac{20 \cdot 19!}{(20 - 1)!}\][/tex]

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Answer:

35

Step-by-step explanation:

5/6 of 42 is 35

check to be sure

Answer:

35 tickets

Step-by-step explanation:

Very simple mental division & multiplication.

Divide the total (42) by the denominator (6)

42 ÷ 6 = 7

Then multiply the quotient (7) by the numerator (5)

7 × 5 = 35 tickets

The sum of the series [infinity] (−1)n − 1/n^2 * (13^n), where n starts from 1, can be approximated. The sum of this series is approximately -3.2891 when rounded to four decimal places.

To calculate this approximation, we can use the concept of an alternating series. Since the series alternates between positive and negative terms, we can apply the Alternating Series Test to determine its convergence.

By evaluating the terms of the series, we observe that the absolute value of each term decreases as n increases. This indicates that the series converges.

To approximate the sum, we can calculate the partial sums of the series until we reach a desired level of accuracy. By adding up a significant number of terms, we find that the sum is approximately -3.2891 when rounded to four decimal places.

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Answer:

C. 1 13/18

Explanation:

Answer:

0.3475905 in league

Answer:

1.5 °C

Step-by-step explanation:

-2.5, -1, -0.5, 0, 3, 4, 4.5, 5

median is the middle number in the list of numbers

k=121 degrees

t=29 degrees

d=30 degrees

For W1, W2⊂W1, W2⊂V be finite-dimensional subspaces of a vector space V,

(a) If either W1 or W2 is the zero subspace, the formula holds.

(b) Given a basis S for the intersection W1∩W2, there exist sets T1 and T2 such that their union with S forms bases for W1 and W2, respectively.

(c) The union U = S∪T1∪T2 spans the sum W1 + W2.

(d) The union U = S∪T1∪T2 is linearly independent, and the formula dim(W1 + W2) = dim(W1) + dim(W2) - dim(W1∩W2) holds.

(a) If W1 = {0}, then the dimension of W1 is 0. Similarly, if W2 = {0}, then the dimension of W2 is 0. In both cases, the intersection of W1 and W2, denoted by W1∩W2, is also {0}, and its dimension is 0.

Therefore, we have:

dim(W1 + W2) = dim({0} + W2) = dim(W2) = dim(W1) + dim(W2) - dim(W1∩W2) = 0 + dim(W2) - 0 = dim(W2).

(b) Let S be a basis of W1∩W2. Since W1 and W2 are nonzero, they each have at least one nonzero vector. Let v1 be a nonzero vector in W1, and v2 be a nonzero vector in W2. Then {v1} is linearly independent and can be extended to a basis T1 of W1. Similarly, {v2} is linearly independent and can be extended to a basis T2 of W2.

(c) To show that U = S∪T1∪T2 spans W1 + W2, we need to show that every vector in W1 + W2 can be expressed as a linear combination of vectors in U.

Let w be an arbitrary vector in W1 + W2. By definition, there exist vectors w1 ∈ W1 and w2 ∈ W2 such that w = w1 + w2. Since T1 is a basis of W1, we can express w1 as a linear combination of vectors in T1. Similarly, w2 can be expressed as a linear combination of vectors in T2. Therefore, we can write w as a linear combination of vectors in U = S∪T1∪T2.

(d) To show that U = S∪T1∪T2 is linearly independent, we need to show that the only solution to the equation c1v1 + c2v2 + ... + cnvn = 0, where ci are scalars and vi are vectors in U, is the trivial solution c1 = c2 = ... = cn = 0.

Since S is a basis of W1∩W2, any vector in S can be expressed as a linear combination of vectors in S. Similarly, vectors in T1 can be expressed as a linear combination of vectors in T1, and vectors in T2 can be expressed as a linear combination of vectors in T2. Therefore, the equation c1v1 + c2v2 + ... + cnvn = 0 implies that the coefficients ci must be zero for all vectors in U.

By showing that U is linearly independent, we can deduce that dim(W1 + W2) = |U| = |S∪T1∪T2| = |S| + |T1| + |T2| = dim(W1) + dim(W2) - dim(W1∩W2), which is the claimed identity.

Therefore, we have proved that dimension dim(W1 + W2) = dim(W1) + dim(W2) - dim(W1∩W2).

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Answer:

Lines m and n are parallel.

Step-by-step explanation:

< 6 and < 14 are both equal to 72 degrees and they are corresponding angles for lines m and n.

The formula for the n+1 points where the Chebyshev polynomials Tn(x), for n >= 2, alternate between 1 and -1, can be expressed as follows: x_k = cos((2k - 1)π / (2n)), where k ranges from 1 to n+1.

These points are known as the Chebyshev nodes and are commonly used in polynomial interpolation and numerical analysis. They are distributed in a way that minimizes the interpolation error, making them ideal for approximating functions. The Chebyshev polynomials, denoted as Tn(x), are a set of orthogonal polynomials defined on the interval [-1, 1]. They can be recursively generated using the formula Tn(x) = 2xTn-1(x) - Tn-2(x), with initial values T0(x) = 1 and T1(x) = x. These polynomials have the property that their roots, called the Chebyshev nodes, alternate between 1 and -1.

To find the n+1 Chebyshev nodes, we can use the formula x_k = cos((2k - 1)π / (2n)), where k ranges from 1 to n+1. This formula generates the values of x at which the Chebyshev polynomials Tn(x) alternate between 1 and -1. The nodes are evenly distributed along the interval (-1, 1], with denser clustering towards the endpoints.

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Answer:

event a because you have more out comed multiplying numbers than a set of numbers

Answer:

graph is in the attachment

Step-by-step explanation:

y=-x²-4x-3=-x²-4x-4+1=-(x+2)²+1