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3 years ago

I need help on outdated I-ready! -.-

Mathematics

1 answer:

tiny-mole [99]3 years ago

4 0

S equals 36

multiply both sides of the equation by 3

S+42=78

Move constant to the right side and change its sign

S=78-42

subtract the numbers

S=36

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The lengths of pregnancies are normally distributed with a mean of days and a standard deviation of days. a. Find the probabilit

Alik [6]

Answer:

a) The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b) We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

Step-by-step explanation:

Normal Probability Distribution

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

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

In this question:

Mean $\mu$ , standard deviation $\sigma$

a. Find the probability of a pregnancy lasting X days or longer.

The probability of a pregnancy lasting X days or longer is given by 1 subtracted by the p-value of $Z = \frac{X - \mu}{\sigma}$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

b. If the length of pregnancy is in the lowest a%, then the baby is premature. Find the length that separates premature babies from those who are not premature.

We have to find X when Z has a p-value of $\frac{a}{100}$ , and X is given by: $X = \mu - Z\sigma$ , in which $\mu$ is the mean and $\sigma$ is the standard deviation.

8 0

3 years ago

What is the converse of the statement below?<br> “If today is Friday, then tomorrow is Saturday."

Ugo [173]

Answer:

C (Sorry if its incorrect)

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A cell phone company is looking to place a new tower to service 3 cities. In order for it to best service each city, it needs to

AlladinOne [14]

A circle is a collection of points all of which are on the circumference of the circle. If you use algebra and (x-h)^2 + (y-k)^2 = r^2 and the info contained in the coordinates (x,y) of each city, you can find h, k and r. Then r is the (equal) distance between each city and the center of the circle, where the new tower will be built.

3 0

3 years ago

Quadrilateral P Q R S is shown. Diagonals are drawn from point Q to point S and from point P to point R and intersect at point M

kobusy [5.1K]

Your anwser for your question is QP Qr and Qrs

7 0

2 years ago

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Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)

USPshnik [31]

See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
$M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) + f(\frac{x_{1} + x_{2} }{2}) + ... + f(\frac{x_{n-1} + x_{n} }{2})]$

In your case:
a = 0
b = 1
n = 4
x₀ = 0
x₁ = 1/4
x₂ = 1/2
x₃ = 3/4
x₄ = 1

Therefore, you'll have:
$M_{4} = \frac{1 - 0}{4} [ f(\frac{0 + \frac{1}{4} }{2}) + f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) + f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]$
$M_{4} = \frac{1}{4} [ f(\frac{1}{8}) + f(\frac{3}{8}) + f(\frac{5}{8}) + f(\frac{7}{8})]$

Now, to evaluate your f(x), you need to look at the graph and notice that:
f(x) = x - x³

Therefore:
$M_{4} = \frac{1}{4} [(\frac{1}{8} - (\frac{1}{8})^{3}) + (\frac{3}{8} - (\frac{3}{8})^{3}) + (\frac{5}{8} - (\frac{5}{8})^{3}) + (\frac{7}{8} - (\frac{7}{8})^{3})]$

$M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]$

M₄ = 1/4 · (2 - 478/512)
= 0.2666

Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

6 0

3 years ago