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Law Incorporation [45]
3 years ago
15

What value of x and y will make the polygon a parallelogram 105° V 15x

Mathematics
1 answer:
Maslowich3 years ago
8 0

Answer

60

Step-by-step explanation:

105+15= 120

180-120=60

I think this is the correct answer but I dont know

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Please answer this quickly !
nika2105 [10]

Answer:

The first option.

Because we're solving for the price of each gallon, x will represent the cost of each gallon. Since 8.4 is the total amount of gallons, multiply 8.4 and x to get $23.94. Solve for x by dividing 23.94 to get $2.85 per gallon. To check this, substitute 2.85 for x into the equation: 8.4 (2.85) = 23.94

4 0
3 years ago
Luke invited 20 more boys than girls to his birthday party. However, only ¾ of the boys and ⅔ of the girls came. Exactly 19 of L
sveta [45]

Answer:

68

Step-by-step explanation:

Let:

Number of girls invited = x

Number of boys = x +20

3/4 of boy showed up

2/3 of girls showed up

Exactly 19 did not show up

Therefore, we could say:

Fraction of boys that didn't show up = 1 - 3/4 = 1/4

Fraction of girls that didn't show up = 1 - 2/3 = 1/3

That is,

[1/4 *(x + 20)] + [1/3*x] = 19

[(x + 20) / 4] + x /3 = 19

(3(x + 20) + 4x) / 12 = 19

3x + 60 + 4x = 228

7x = 228 - 60

7x = 168

x = 168 / 7

x = 24

Number of girls, x = 24

Number of boys, x + 20 = 24 + 20 = 44

Total friends invited = (24 + 44) = 68

8 0
3 years ago
 What is 4/1 + 3/4 =  equal to?
kari74 [83]
4/1 is the same as 4.

So 4+3/4= 4 and 3/4

<u>Final Answer</u>= 4 and 3/4
3 0
3 years ago
Read 2 more answers
Compute the following volume and surface area.
Sidana [21]
Isssssssssssssssssssssssssssss: 192...............................................that is the correct ansewer aaaaa.........A
5 0
3 years ago
Read 2 more answers
In a large lecture class, the professor announced that the scores on a recent exam were normally distributed with a range from 5
iVinArrow [24]

Answer:

The standard deviation of the scores on a recent exam is 6.

The sample size required is 25.

Step-by-step explanation:

Let <em>X</em> = scores of students on a recent exam.

It is provided that the random variable <em>X</em> is normally distributed.

According to the Empirical rule, 99.7% of the normal distribution is contained in the range, <em>μ </em>± 3<em>σ</em>.

That is, P (<em>μ </em>- 3<em>σ </em>< <em>X</em> < <em>μ </em>+ 3<em>σ</em>) = 0.997.

It is provided that the scores on a recent exam were normally distributed with a range from 51 to 87.

This implies that:

P (51 < <em>X</em> < 87) = 0.997

So,

<em>μ </em>- 3<em>σ </em>= 51...(i)

<em>μ </em>+ 3<em>σ </em>= 87...(ii)

Subtract (i) and (ii) to compute the value of <em>σ</em> as follows:

<em>   μ </em>-     3<em>σ </em>=    51

(-)<em>μ </em>+ (-)3<em>σ </em>= (-)87

______________

-6<em>σ </em>= -36

<em>σ</em> = 6

Thus, the standard deviation of the scores on a recent exam is 6.

The (1 - <em>α</em>)% confidence interval for population mean is given by:

CI=\bar x\pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

The margin of error of this interval is:

MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

Given:

MOE = 2

<em>σ</em> = 6

Confidence level = 90%

Compute the <em>z</em>-score for 90% confidence level as follows:

z_{\alpha/2}=z_{0.10/2}=z_{0.05}=1.645

*Use a <em>z</em>-table.

Compute the sample required as follows:

MOE = z_{\alpha/2}\frac{\sigma}{\sqrt{n}}\\2=1.645\times \frac{6}{\sqrt{n}}\\n=(\frac{1.645\times 6}{2})^{2}\\n=24.354225\\n\approx 25

Thus, the sample size required is 25.

4 0
3 years ago
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