Ask question

Ask question

All categories

1 year ago

8

the times of all 15 year olds who run a certain race are approximately normally distributed with a given mean

1 answer:

Butoxors [25]1 year ago

8 0

The percentage of the runners who have times less than 14.4 sec is 0.15%

The times of all 15-year old who run a certain race are approximately normally distributed with a given mean of 18 sec and a standard deviation of 1.2 sec.

We know that,

where,

X = raw score = 14.4

μ = mean = 18

σ = standard deviation = 1.2

Putting the values,

Finding the probability from the z score table, we get

P(z < -3) = 0.00135 = 0.135%

Disclaimer: The question is incomplete. Please read below to find the missing content.

Question: The times of all 15-year-olds who run a certain race are approximately normally distributed with a given mean of 026-1 = 18 seconds and a standard deviation of 026-2. = 1.2 sec. What percentage of the runners have times less than 14.4 seconds?

0.15%

0.15%

0.30%

0.60%

2.50%

Learn more about percentages here brainly.com/question/809966

#SPJ4

You might be interested in

Does anyone know algebra? if so, I can really use your help.

SCORPION-xisa [38]

Answer:

yea sure

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

Find the missing angle measure A=51°, B=82°, C=5x+2

Shalnov [3]

I'm assuming you mean triangle angles?
If so, all angles add up to 180 so combine them and set them equal to 180.
51+82+5x+2=180
135+5x=180
5x=45
x=9

8 0

3 years ago

Please Help Me!!!!!!!!!!!

Tju [1.3M]

Answer:

Undefined

Step-by-step explanation:

This is horizontal line, so the slope is undefined.

4 0

3 years ago

Read 2 more answers

Which option lists am expression that is not equivalent to 4^2/3

Vesnalui [34]

remember that

$x^{\frac{a}{b}}=\sqrt[b]{x^a}$

also $x^{-a}=\frac{1}{x^a}$

and $(a^b)^c=a^{bc}$

so

$4^{\frac{2}{3}}$

first one, that one is clearly not equal since 0.25≠4

2nd one, 0.25=1/4, so $0.25^{\frac{-2}{3}}=(\frac{1}{4})^{\frac{-2}{3}=$ $\frac{1}{(\frac{1}{4})^{\frac{2}{3}}}=4^{\frac{2}{3}}$ , which matches

3rd one, $\sqrt[3]{16}=\sqrt[3]{4^2}=4^{\frac{3}{2}}$ , which matches

4th one $(\sqrt[3]{4})^2=(4^{\frac{1}{3}})^2=4^{\frac{2}{3}}$ which matches

answer is $0.25^{\frac{2}{3}}$

4 0

3 years ago

119. Beatrice decides to deposit $100 per month at the end of every month in a bank with an annual interest rate of 5.5% compoun

sveta [45]

Answer:

Beatrice will accumulate $1230.72 at the end of the year.

Step-by-step explanation:

We can write:

$(1+\frac{0.055}{12})=z$

$100=C$ for deposits

The first month would have only the deposit reflected in her balance, then, expanding some steps of the calculation would yield:

$S_{1}=C\\S_{2}=Cz + C\\S_{3}=Cz^{2}+Cz + C\\S_{n}=Cz^{n-1} +...+Cz+C$

A geometric series is given by:

$1+x+...+x^{m}=\frac{x^{m+1}-1}{x-1}$

Translating our series to the short form:

$S_{n}=C(\frac{z^{n}-1}{z-1} )$

plugin in the values for the 12 month gives:

$S_{12}=100(\frac{(1+\frac{0.055}{12})^{12} -1}{\frac{0.055}{12}})=1230.7169$

3 0

3 years ago