A rectangle has an area of 54 square inches. The width of the rectangle is 9 inches. What is the length of the rectangle?

Helppppppp pleaseeeeee

Dahasolnce [82]

Step-by-step explanation:

The area, A, in square yards, of a rectangular garden, is given by the function.

$A(x)=-2x^2+30x$

We know that, the area of the rectangle is given by :

$A=l\times b$

Where l is length and b is breadth

$A=-2x^2+30x\\\\A=2x\times (-x+15)$

Length = 2x

Breadth = (15-x)

For zeoes,

$2x\times (-x+15)=0\\\\x=15$

Length = 15 yards

Breadth = (15-15) = 0 yards

Hence, this is the required solution.

4 0

3 years ago

A machine that fills beverage cans is supposed to put 12 ounces of beverage in each can. The standard deviation of the amount in

erica [24]

Answer:

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

Step-by-step explanation:

Assuming the following data:"12.14 12.05 12.27 11.89 12.06

12.14 12.05 12.38 11.92 12.14"

We can calculate the sample deviation with this formula:

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$n=10$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =0.0217$ represent the sample variance

$\sigma^2_0 =0.12^2= 0.0144$ represent the value to verify

Null and alternative hypothesis

We want to determine whether the standard deviation differs from 0.12 ounce, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 = 0.0144$

Alternative hypothesis: $\sigma^2 \neq 0.0144$

The statistic can be calculated like this;

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

$\chi^2 =\frac{10-1}{0.0144} 0.0217 =13.54$

The degrees of freedom are:

$df=n-1=10-1=9$

Now we can calculate the p value using the alternative hypothesis:

$p_v =2*P(\chi^2 >13.54)=0.279$

Since the p value is higher than the signficance level assumed of 0.05 we have enough evidence to FAIL to reject the null hypothesis and there is no evidence to conclude that the true deviation differs from 0.12 ounces

7 0

3 years ago

HELP PLZ <br> i need help with this math side i just need b,c,d a =4x+2y=16

tino4ka555 [31]

Answer:

Please check the explanatio.

Step-by-step explanation:

Solving Part b)

We know that the slope-intercept form of the equation line is

$y=mx+b$