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3 years ago

The two triangles are similar. What is the value of x?

Mathematics

1 answer:

denpristay [2]3 years ago

3 0

[8/(2x-2)=(8+6)/(3x)
8/(2x-2)=14/(3x)
14(2x-2)=8(3x)
28x-28=24x
x=7
So by using the similarity of the sides, we can make ratios that can give is the value of x.<span />

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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

What is the value of the digit7in the population of Memphis

Keith_Richards [23]

70,00 Is the the answer

7 0

3 years ago

Hi im so good at algebra i can replace your x and you wouldnt even know y

Naddika [18.5K]

Answer:

If you were words on a page, you'd be <em>Fine Print.</em>

6 0

3 years ago

If angle a has the terminal ray that falls in the fourth quadrant and cosine a equals 5/9 then determine the value of sin a in s

Tema [17]

Answer: $-\dfrac{2\sqrt{14}}{9}$

Step-by-step explanation:

Given

$\cos a=\dfrac{5}{9}$

$a$ lies in the fourth quadrant

So, sine must be negative in the fourth quadrant

Using identity $\sin ^2 x+\cos^2 x=1$ to find sine value

$\Rightarrow \sin^2 a=1-\dfrac{5^2}{9^2}$

$\\\\\Rightarrow \sin^2 a=1-\dfrac{25}{81}\\\\\\\Rightarrow \sin^2 a=\dfrac{56}{81}\\\\\\\Rightarrow \sin a=-\sqrt{\dfrac{56}{81}}\\\\\\\Rightarrow \sin a=-\dfrac{2\sqrt{14}}{9}$

8 0

2 years ago

Shawn found that a solution of x = 3 was extraneous for his function. What would be a possible denominator in Shawn's function?

Alex_Xolod [135]

Your answer is B. x2 -2x -3

Hope this helps!!

6 0

2 years ago

Read 2 more answers