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arsen [322]
3 years ago
5

How do you do this ??

Mathematics
1 answer:
VikaD [51]3 years ago
7 0

Answer:

(D)

Step-by-step explanation:

First, you have to graph the equation. Then you notice that it is a parabola facing downwards. This means that as the graph decreases on the left side, the x's become negative and the y's become negative. On the right side, the x's become positive and the y's become negative.

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In the following problem, check that it is appropriate to use the normal approximation to the binomial. Then use the normal dist
frosja888 [35]

Answer:

a) 0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

b) 0.2946  = 29.46% probability that 30 or more will live beyond their 90th birthday

c) 0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday

d) 0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

Step-by-step explanation:

We solve this question using the normal approximation to the binomial distribution.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

Sample of 723, 3.7% will live past their 90th birthday.

This means that n = 723, p = 0.037.

So for the approximation, we will have:

\mu = E(X) = np = 723*0.037 = 26.751

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{723*0.037*0.963} = 5.08

(a) 15 or more will live beyond their 90th birthday

This is, using continuity correction, P(X \geq 15 - 0.5) = P(X \geq 14.5), which is 1 subtracted by the pvalue of Z when X = 14.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{14.5 - 26.751}{5.08}

Z = -2.41

Z = -2.41 has a pvalue of 0.0080

1 - 0.0080 = 0.9920

0.9920 = 99.20% probability that 15 or more will live beyond their 90th birthday

(b) 30 or more will live beyond their 90th birthday

This is, using continuity correction, P(X \geq 30 - 0.5) = P(X \geq 29.5), which is 1 subtracted by the pvalue of Z when X = 29.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{29.5 - 26.751}{5.08}

Z = 0.54

Z = 0.54 has a pvalue of 0.7054

1 - 0.7054 = 0.2946

0.2946  = 29.46% probability that 30 or more will live beyond their 90th birthday

(c) between 25 and 35 will live beyond their 90th birthday

This is, using continuity correction, P(25 - 0.5 \leq X \leq 35 + 0.5) = P(X 24.5 \leq X \leq 35.5), which is the pvalue of Z when X = 35.5 subtracted by the pvalue of Z when X = 24.5. So

X = 35.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{35.5 - 26.751}{5.08}

Z = 1.72

Z = 1.72 has a pvalue of 0.9573

X = 24.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{24.5 - 26.751}{5.08}

Z = -0.44

Z = -0.44 has a pvalue of 0.3300

0.9573 - 0.3300 = 0.6273

0.6273 = 62.73% probability that between 25 and 35 will live beyond their 90th birthday.

(d) more than 40 will live beyond their 90th birthday

This is, using continuity correction, P(X > 40+0.5) = P(X > 40.5), which is 1 subtracted by the pvalue of Z when X = 40.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{40.5 - 26.751}{5.08}

Z = 2.71

Z = 2.71 has a pvalue of 0.9966

1 - 0.9966 = 0.0034

0.0034 = 0.34% probability that more than 40 will live beyond their 90th birthday

6 0
3 years ago
PLZ PLZ HURRY I WILL GIVE BRIANLEST
Colt1911 [192]
The answer is D Explanation: since the the slope is negative the x should be negative and D is the only one that has a negative X.
8 0
3 years ago
Read 2 more answers
If A=[xy5−8] determine the values of x and y for which A2=A.
Ivanshal [37]

Answer:

<h2>x = 9, y = -72/5</h2>

Step-by-step explanation:

Given the 2 by 2 matrix A = \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right] , we are to find the value of x and y for the expression A² = A to be true.

First we need to find A² by multiplying the same matrix together

A^2 = \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right] \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right]\\\\ \ we \ normally \ multiply \ the \ rows \ of \ the \ first \ matrix \ with \ the \ column \ of \ the \ second \\\\A^2 = \left[\begin{array}{-cc}x^2+5y&5x-40\\xy-8y&5y+64\\\end{array}\right]\\\\Since \ A^2 = A,\ hence;\\\\\left[\begin{array}{-cc}x^2+5y&5x-40\\xy-8y&5y+64\\\end{array}\right] =  \left[\begin{array}{-cc}x&5\\y&-8\\\end{array}\right]\\

Equating the first row and second column of both matrices together, we will have;

5x-40 = 5

add 40 to both sides of the equation

5x-40+40 = 5+40

5x = 45

x = 45/5

x = 9

Similarly,  we will equate the second row and second column of both matrices to have;

5y+64 = -8

Subtract 64 from both sdies

5y+64-64 = -8-64

5y = -72

y = -72/5

<em>Hence the value of x is 9 and y is -72/5</em>

7 0
4 years ago
Find the value of given expression<br><br><img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B4%7D%20%5Ctimes%20%20%282%29" id="TexFo
Neko [114]

Step-by-step explanation:

Given

[tex] \sqrt{4} \times (2) \\ \sqrt{ {2}^{2} } \times 2 \\ = 2 \times 2 \\ = 4

Hope it will help :)❤

8 0
3 years ago
Read 2 more answers
Solve the equation x-23=18​
Nastasia [14]

Answer:

x = 41

Step-by-step explanation:

x-23=18​

Add 23 to each side

x-23+23=18+23

x =41​

4 0
3 years ago
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