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2 years ago

What is the value of x? Enter your answer, as a decimal, in numerical value only.

Mathematics

1 answer:

Doss [256]2 years ago

7 0

Answer:

x = 67.5

Step-by-step explanation:

If a line is parallel to a side of a triangle and intersects the other 2 sides , it divides those sides proportionally.

AB is parallel to NP and intersects the other 2 sides , then

$\frac{MA}{AN}$ = $\frac{MB}{BP}$ ( substitute values )

$\frac{71.5-22}{22}$ = $\frac{x}{97.5-x}$ , that is

$\frac{49.5}{22}$ = $\frac{x}{97.5-x}$ ( cross- multiply )

49.5(97.5 - x) = 22x ← distribute left side

4826.25 - 49.5x = 22x ( add 49.5x to both sides )

4826.25 = 71.5x ( divide both sides by 71.5 )

67.5 = x

Read more about equations at:

brainly.com/question/15349799

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<u>Complete question</u>

Which equation has infinite solutions?

3(x – 1) = x + 2(x + 1) + 1

x – 4(x + 1) = –3(x + 1) + 1

$2x + 3 = \frac{1}{2}(4x + 2) + 2$

$\frac 13(6x - 3) = 3(x + 1) - x - 2$

5 0

2 years ago

A random sample of 20 students yielded a mean of ¯x = 72 and a variance of s2 = 16 for scores on a college placement test in mat

Helen [10]

Answer:

The 98% confidence interval for the variance in the pounds of impurities would be $8.400 \leq \sigma^2 \leq 39.827$ .

Step-by-step explanation:

1) Data given and notation

$s^2 =16$ represent the sample variance

s=4 represent the sample standard deviation

$\bar x$ represent the sample mean

n=20 the sample size

Confidence=98% or 0.98

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population mean or variance lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

The Chi square distribution is the distribution of the sum of squared standard normal deviates .

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

$\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}$

On this case the sample variance is given and for the sample deviation is just the square root of the sample variance.

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

$df=n-1=20-1=19$

Since the Confidence is 0.98 or 98%, the value of $\alpha=0.02$ and $\alpha/2 =0.01$ , and we can use excel, a calculator or a table to find the critical values.