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Rasek [7]
3 years ago
6

14 = 2y/5 what is the solution?

Mathematics
1 answer:
Gnom [1K]3 years ago
5 0

Answer:

y=35

Step-by-step explanation:

y in (-oo:+oo)

14 = (2*y)/5 // - (2*y)/5

14-((2*y)/5) = 0

(-2/5)*y+14 = 0

14-2/5*y = 0 // - 14

-2/5*y = -14 // : -2/5

y = -14/(-2/5)

y = 35

y = 35

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3 years ago
Here are the ingredients needed to make 8 pancakes.
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Answer:

a) 1000mL Milk

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Step-by-step explanation:

a) 250mL*4=1000mL

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8 0
3 years ago
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Because of safety considerations, in May 2003 the Federal Aviation Administration (FAA) changed its guidelines for how small com
Rasek [7]

Answer:

a) The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

b)

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c) They are respected, as the upper bound of both intervals is below the new FAA recommendations.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve these questions.

Question a:

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of 1 - \frac{1 - 0.95}{2} = 0.975. So we have T = 1.9842

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{20}{\sqrt{100}} = 4

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 183 - 4 = 179 pounds.

The upper end of the interval is the sample mean added to M. So it is 183 + 4 = 187 pounds.

The 95% confidence interval for the mean summer weight (including carry-on luggage) of Frontier Airlines passengers is between 179 and 187 pounds. This means that we are 95% sure that the mean summer weight of all Frontier Airlines passengers is between these two values.

Question b:

Critical value is the same(same sample size and confidence level).

The margin of error is:

M = T\frac{s}{\sqrt{n}} = 1.9842\frac{23}{\sqrt{100}} = 4.6

The lower end of the interval is the sample mean subtracted by M. So it is 190 - 4.6 = 185.4 pounds.

The upper end of the interval is the sample mean added to M. So it is 190 + 4.6 = 194.6 pounds.

The 95% confidence interval for the mean winter weight (including carry-on luggage) of Frontier Airlines passengers is between 185.4 pounds and 194.6 pounds. This means that we are 95% sure that the mean winter weight of all Frontier Airlines passengers is between these two values.

c. The new FAA recommendations are 190 pounds for summer and 195 pounds for winter. Comment on these recommendations in light of the confidence interval estimates from Parts (a) and (b).

They are respected, as the upper bound of both intervals is below the new FAA recommendations.

7 0
3 years ago
If y = 9.99 inches, what is the perimeter of the rhombus shown above
Finger [1]

Answer:

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zavuch27 [327]
Your numbers are .17 and .165 meters

Let me explain each digit value:

Lets start with .17

Decimals are like whole numbers where each placement has a digit value

.1 is the tenths place

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.1 is the tenths place

.06 is the hundredths place

.005 is the thousandths place

When seeing which decimal is larger, go along the tenths place and move along.


So as you can see
, .17 and .165 start with .1 so you can't see which is bigger by this. Now move on to the next digit. Which is larger? .17 or .16? As you should know, 7 is bigger than 6 so that means .17 is larger

Answer: .17 meters is larger than .165 meters
3 0
3 years ago
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