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2 years ago

What is 4(7k + ( -2a) + 3a + 4)

Mathematics

1 answer:

kogti [31]2 years ago

4 0

Answer:

Step-by-step explanation:

(4) (7k+-2a+3a+4)

=(4) (7k) +(4) (-2a) +(4) (3a) +(4)

=28k-8a+12a+16

=4a+28k+16

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*What is the perimeter of a square with side length 13.4 inches? Write the direct variation equation.

Zarrin [17]

Answer:

53.6 inches

p = 4s

Step-by-step explanation:

The perimeter of a square is four times the side

The direct variation equation is

p = 4s

p = 4 * 13.4

p = 53.6 inches

5 0

2 years ago

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

247x82 using in area model

vovikov84 [41]

247x82=534 5 go in the hundreads place 3 go in the tens place and 4 go in the ones place

8 0

3 years ago

Use the equation n= b-4 to find the value of n when b=9

Valentin [98]

Put 9 in for b

N=9-4

N=5

6 0

3 years ago

solve systems of equation. equation number 1: 9 x squared + 4 y squared equals 144. equation 2: x squared plus y squared equals

grigory [225]

$9x^{2} + 4y^{2} =144$
$x^{2} + y^{2}=24$

$9x^{2} + 4y^{2} =144$
$4x^{2} + 4y^{2}=96$
by the difference
5 $x^{2}$ =48
so x= $\sqrt{\frac{48}{5}}$ or x=- $\sqrt{\frac{48}{5}}$
and $y^{2}$ =24- $\frac{48}{5}$
then y= $\sqrt{\frac{72}{5}}$ or - $\sqrt{\frac{48}{5}}$

5 0

3 years ago