Seven less than nine times a number is negative seven. What is the number?

A rope is 50 meters long. It is cut into two pieces and one piece is 8 meters longer than the other. What are the two lengths?

Mashcka [7]

Answer:

21 meters and 29 meters

Step-by-step explanation:

x + (x + 8) = 50

2x + 8 = 50

2xc = 42

x = 21

x + 8 = 29

6 0

3 years ago

Solving rational equations solve for x: 3/x-4=7/x

KIM [24]

Solving the rational equation $\frac{3}{x-4}=\frac{7}{x}$ the value of x is 7

Step-by-step explanation:

We need to solve rational equation: $\frac{3}{x-4}=\frac{7}{x}$ and find value of x.

Solving:

$\frac{3}{x-4}=\frac{7}{x}$

Cross multiplying:

$3(x)=7(x-4)\\Simplifying:\\3x=7x-28\\3x-7x=-28\\-4x=-28\\x=-28/-4\\x=7$

So, value of x = 7.

Solving the rational equation $\frac{3}{x-4}=\frac{7}{x}$ the value of x is 7

Keywords: Solving the rational equation

Learn more about Solving the rational equation at:

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8 0

3 years ago

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The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95