How do I solve this?

Hey do any of yall now what 65'6 x777

katrin2010 [14]

Answer:

If you mean 656, then the answer is 509,712.

Edit: If you're talking about measurements, then it would be 50,893 feet and 6 inches.

6 0

1 year ago

Find the equation of the line that is perpendicular to the x-axis and passes through the point (14,-9). Give the

Rudiy27

Answer:

y = -9

Step-by-step explanation:

a line perpendicular to the x-axis is a horizontal line (which means the slope is zero) which pass through the y-axis at -9

3 0

2 years ago

The difference of twice a number and 4 is 16 .

german

Answer:

Let X be the number.

Twice the number would be 2x

The difference between the two, would be subtraction, so you would subtract 4 from 2x

The equation becomes 2x-4 = 16

Now solve for x:

2x-4 = 16

Add 4 to each side:

2x = 20

Divide both sides by 2:

x = 20/2

x = 10

The number is 10.

7 0

3 years ago

Read 2 more answers

Simplify 14/70<br>please show work

Mumz [18]

Cancel the common factors in the numerator and denominator.

Exact Form:

15

Decimal Form:

0.2

hope l helped have a good day

3 0

3 years ago

Read 2 more answers

A survey of 76 commercial airline flights of under 2 hours resulted in a sample average late time for a flight of 2.55 minutes.

nika2105 [10]

Answer:

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

Step-by-step explanation:

Information given

$\bar X=2.55$ represent the sample mean for the late time for a flight

$\mu$ population mean

$\sigma=12$ represent the population deviation

n=76 represent the sample size

Confidence interval

The best point of estimate for the true mean is:

$\hat \mu = \bar X = 2.55$

The confidence interval for the true mean is given by:

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)

The Confidence level given is 0.95 or 95%, th significance would be $\alpha=0.05$ and $\alpha/2 =0.025$ . If we look in the normal distribution a quantile that accumulates 0.025 of the area on each tail we got $z_{\alpha/2}=1.96$

Replacing we got:

$2.55-1.96\frac{12}{\sqrt{76}}=-0.148$

$2.55+1.96\frac{12}{\sqrt{76}}=5.248$

Since the time can't be negative a good approximation for the confidence interval would be (0,5.248) minutes. The interval are tellling to us that at 95% of confidence the average late time is lower than 5.248 minutes.

7 0

3 years ago