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loris [4]
3 years ago
7

How do I solve this?

Mathematics
1 answer:
kobusy [5.1K]3 years ago
3 0

Answer:

3q^2/5p^4

Step-by-step explanation:

any variable/# w/a negative exponent is also 1/x^#

ex: p^-2 = 1/p^2 -> 9q^5/15p^4q^3

When dividing, you're subtracting the exponents

ex: q^5/q^3 = q^5-3 -> q^2

Lastly, simplify 9/15 to 3/5 -> 3q^2/5p^4

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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
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