-3(8x+3)-2(1-17x)-3(1+3x)=0

Lengths of full-term babies in the US are Normally distributed with a mean length of 20.5 inches and a standard deviation of 0.9

mash [69]

Answer:

66.48% of full-term babies are between 19 and 21 inches long at birth

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.

Mean length of 20.5 inches and a standard deviation of 0.90 inches.

This means that $\mu = 20.5, \sigma = 0.9$

What percentage of full-term babies are between 19 and 21 inches long at birth?

The proportion is the p-value of Z when X = 21 subtracted by the p-value of Z when X = 19. Then

X = 21

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

$Z = \frac{21 - 20.5}{0.9}$

$Z = 0.56$

$Z = 0.56$ has a p-value of 0.7123

X = 19

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

$Z = \frac{19 - 20.5}{0.9}$

$Z = -1.67$

$Z = -1.67$ has a p-value of 0.0475

0.7123 - 0.0475 = 0.6648

0.6648*100% = 66.48%

66.48% of full-term babies are between 19 and 21 inches long at birth

5 0

2 years ago

Question 1 of 12, Step 1 of 1

stellarik [79]

Answer:

325 and 375 mph

Step-by-step explanation:

Given :

Distance between plane A and B = 2800

Recall :

Distance = speed * time

Let speed of A = v1

Speed of B = v2

v1 = v2 + 50

Time taken = 4 hours

Distance = speed * time

Total distance = 2800

2800 = v1 * 4 + v2 * 4

2800 = (4v1) + (4v2)

2800 = 4(v2+50) + 4v2

2800 = 4v2 + 200 + 4v2

2800 = 8v2 + 200

2800 - 200 = 8v2

2600 = 8v2

v2 = 2600/8

v2 = 325 mph

v1 = v2 + 50

v1 = 325 + 50

v1 = 375 mph

8 0

2 years ago

What is 40 percent of 50

valentinak56 [21]

20 is the answer to the question you asked.

3 0

3 years ago

Read 2 more answers

Can yall help me out

Ainat [17]

Answer:

Option D, Yes, Devon should have evaluated f(3)

Step-by-step explanation:

f(x) = x^3 + x^2 − 10x + 8

<u>Step 1: Evaluate using F(3) NOT f(-3)</u>

f(3) = 3^3 + 3^2 - 10(3) + 8

f(3) = 27 + 9 - 30 + 8

f(3) = 14

Since the factor is (x - 3) the x value is x - 3 + 3 = 0 + 3 -> x = 3 not x = -3

Answer: Option D, Yes, Devon should have evaluated f(3)

6 0

3 years ago

Dose someone know how to figure this out can someone pleas help its a test

riadik2000 [5.3K]

Complete question is;

The levels of mercury in two different bodies of water are rising. In one body of water the initial measure of mercury is 0.05 parts per billion (ppb) and is rising at a rate of 0.1 ppb each year. In the second body of water the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Which equation can be used to find y, the year in which both bodies of water have the same amount of mercury?

A) 0.05 – 0.1y = 0.12 – 0.06y

B) 0.05y + 0.1 = 0.12y + 0.06

C) 0.05 + 0.1y = 0.12 + 0.06y

D) 0.05y – 0.1 = 0.12y – 0.06

Answer:

Option C: 0.05 + 0.1y = 0.12 + 0.06y

Step-by-step explanation:

In the first body, the initial measure of mercury is 0.05 parts per billion (ppb) while it's rising at a rate of 0.1 ppb each year. We are told to use y for the number of years.

Thus, amount of mercury for y years in this first body is;

A1 = 0.05 + 0.1y

Now, for the second body, we are told that;the initial measure is 0.12 ppb and the rate of increase is 0.06 ppb each year.

Thus, amount of mercury for y years in this second body is;

A2 = 0.12 + 0.06y

Since we want to find the year in which both bodies of water have the same amount of mercury. Thus, it means;

A1 = A2

Thus;

0.05 + 0.1y = 0.12 + 0.06y

7 0

2 years ago