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2 years ago

Problem Luke can paint 91portraits in 7 weeks. How many portraits can Luke paint in 4 weeks?

Mathematics

2 answers:

jonny [76]2 years ago

7 0

Divide 91÷7 = 13. 13 is how much in one week. So 13× 4 = 52
52 portrait

ludmilkaskok [199]2 years ago

3 0

91 divided by 7 = 13
4 times 13 equals 52
52 portraits

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A study conducted by the Center for Population Economics at the University of Chicago studied the birth weights of 623 babies bo

Margaret [11]

Answer:

The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

$\mu = 3448, \sigma = 862$

Proportion of newborns who weighed between 1724 grams and 5172 grams.

This is the pvalue of Z when X = 5172 subtracted by the pvalue of Z when X = 1724. So

X = 5172

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

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{5172 - 3448}{862}$

$Z = 2$

$Z = 2$ has a pvalue of 0.9772

X = 1724

$Z = \frac{X - \mu}{s}$

$Z = \frac{1724 - 3448}{862}$

$Z = -2$

$Z = -2$ has a pvalue of 0.0228

0.9772 - 0.0228 = 0.9544

Estimate the number of newborns who weighed between 1724 grams and 5172 grams.

0.9544 out of 623 babies. SO

0.9544*623 = 595

The estimation for the number of newborns who weighed between 1724 grams and 5172 grams is 595.

5 0

3 years ago

Can someone please do this ill mark brainliest

Svet_ta [14]

Answer:

-60 im pretty sure this is it

8 0

3 years ago

Identify the probability to the nearest hundredth that a point chosen randomly inside the rectangle is in the triangle. HELP PLE

sashaice [31]

Answer:

$0.02$

Step-by-step explanation:

we know that

The probability that a point chosen randomly inside the rectangle is in the triangle is equal to divide the area of the triangle by the area of rectangle

Let

x-----> the area of triangle

y----> the area of rectangle

P -----> the probability

$P=x/y$