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3 years ago

Was my formula correct?

Mathematics

1 answer:

Ivanshal [37]3 years ago

3 0

A(b+c)=a*b+a*c CORRECT

2(y+11)=2*y+2*11), BUT better 11y+22 both are CORRET

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Is there any Indian active right now?

Vera_Pavlovna [14]

Answer:

do see the photos

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2 years ago

How do I graph this??? HELP FAST

Luda [366]

Answer:

the 2x and 1/2x are the slope.(rise over run)

the +2 and -1 are intercepts on the y axis

Step-by-step explanation:

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3 years ago

Read 2 more answers

It appears that people who are mildly obese are less active than leaner people. One study looked at the average number of minute

NemiM [27]

Answer:

a) 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

b) 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Normal probability distribution

Problems of normally distributed samples 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.

Mildly obese

Normally distributed with mean 373 minutes and standard deviation 67 minutes. So $\mu = 373, \sigma = 67$

A) What is the probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes?

So $n = 5, s = \frac{67}{\sqrt{5}} = 29.96$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 373}{29.96}$

$Z = 1.23$

$Z = 1.23$ has a pvalue of 0.8907.

So there is a 1-0.8907 = 0.1093 = 10.93% probability that the mean number of minutes of daily activity of the 5 mildly obese people exceeds 420 minutes.

Lean

Normally distributed with mean 526 minutes and standard deviation 107 minutes. So $\mu = 526, \sigma = 107$

B) What is the probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes?

So $n = 5, s = \frac{107}{\sqrt{5}} = 47.86$

This probability is 1 subtracted by the pvalue of Z when X = 410.

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

$Z = \frac{410 - 526}{47.86}$

$Z = -2.42$

$Z = -2.42$ has a pvalue of 0.0078.

So there is a 1-0.0078 = 0.9922 = 99.22% probability that the mean number of minutes of daily activity of the 5 lean people exceeds 420 minutes.

7 0

3 years ago

42 divided by 4 and 67 divided by8

Bezzdna [24]

10.5 and 8.4 ......................,,,,,

6 0

3 years ago

Read 2 more answers

5. The internal dimension of a tank are 14m, 8m and 5m. How much water can be

EleoNora [17]

Answer:

560,000L

Step-by-step explanation:

Given data

L=14m

W=8m

H=5m

Volume= L*W*H

Volume= 14*8*5

Volume= 560 m^3

1m^3 = 1000L

560m^3 = x

cross multiply

x= 560*1000

x= 560,000L

Hence the volume is 560,000L

3 0

2 years ago