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2 years ago

A 17.2 gallon gasoline tank is 1/2 full. How many gallons will it take to fill the tank?

Mathematics

1 answer:

love history [14]2 years ago

3 0

Answer:

8.6

Step-by-step explanation:

We know that the tank is filled halfway, and it can hold 17.2 gallons, so we can just divide to figure out how much is half the tank (the amount we need to fill it)

17.2 / 2 = 8.6

We can check our work by multiplying

8.6 x 2 = 17.2

So it would take 8.6gal to fill it up. I hope this helps :)

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How do you find the area of this

zlopas [31]

Answer:

2345 units^2

Step-by-step explanation:

First, you can use Soh Cah Toa to find line c to be 36*sqrt(2). Furthermore, you can use the same equation to find the other sides to be both 65.15. To further solve, you just need to multiply 65.15 by 36 to get a final answer of 2,345.4 units^2

Please Mark Brainliest if this helped!

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3 years ago

Can someone please check if this is correct?

Solnce55 [7]

Answer:

Just two corrections! See attached image.

Step-by-step explanation:

The product of 8 and b taken from 10 means begin with 10 and subtract 8b from it.

8 times the difference of 10 and b means find 10 - b first, then multiply by 8.

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2 years ago

What is the range of y = sinx in the interval - π ≤ x ≤ 0?

Kaylis [27]

Answer:

$-1 \le y \le 1$

Step-by-step explanation:

Let be $y = \sin x$ , the range of the function corresponds to the set of values of $y$ . The sine function is a bounded function between -1 and 1. Hence, the range of the function is:

$-1 \le y \le 1$

6 0

2 years ago

In a study of the effect of college student employment on academic performance, the following summary statistics for GPA were re

11Alexandr11 [23.1K]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. Let μemployed(μ1) be the sample mean of students who are employed and μnot employed(μ2) be the sample mean of students who are not employed

The random variable is μ1 - μ2 = difference in the mean of the employed and unemployed students.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 < μ2 H1 : μ1 - μ2 < 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 3.22

μ2 = 3.33

s1 = 0.475

s2 = 0.524

n1 = 172

n2 = 116

t = (3.22 - 3.33)/√(0.475²/172 + 0.524²/116)

t = - 1.81

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [0.475²/172 + 0.524²/116]²/[(1/172 - 1)(0.475²/172)² + (1/116 - 1)(0.524²/116)²] = 0.00001353363/0.00000005878

df = 230

We would determine the probability value from the t test calculator. It becomes

p value = 0.036

Since alpha, 0.05 > than the p value, 0.036, then we would reject the null hypothesis.

Therefore, at a 5% significant level, this information support the hypothesis that for students at this university, those who are not employed have a higher mean GPA than those who are employed

3 0

3 years ago

Last year Baron Enterprises had $800 million of sales, and it had $270 million of fixed assets that were used at 65% of capacity

Svetllana [295]

Answer: B. $430.80

Step-by-step explanation:

Given : Last year Baron Enterprises had $800 million of sales.

It had $270 million of fixed assets that were used at 65% (=0.65) of capacity last year.

Now, the used asset = $0.65\times270=\$175.5$ million

Now, Baron Enterprises had $800 million of sales in $175.5 million of assets , if we use all of $270 million of fixed assets , then the sales will be :-

$\dfrac{800}{175.5}\times270=1230.76923077\approx1230.80\text{ million}$

Now, the increase in Baron's sales before it is required to increase its fixed assets = $\$1230.80-$800=\$430.80\text{ millions}$

Hence, the increase in Baron's ( in million ) sales before it is required to increase its fixed assets = $430.80

8 0

3 years ago