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3 years ago

“Write an equation and solve for the unknown” I know it’s a lot but can someone help me with this

Mathematics

1 answer:

Svetradugi [14.3K]3 years ago

5 0

Answer:

g) x = 30

h) x = 30

i) m = 55

j) x = 24

k) k = 45

Step-by-step explanation:

all angles of a triangle add up to 180.

the square in the bottom right of the triangle is 90 btw

2x + x + 90 = 180

3x + 90 = 180

3x + 90 - 90 = 180 - 90

3x = 180 - 90

3x = 90

3x/3 = 90/3

x = 30

this is an Isosceles triangle.

that means the bottom two angles are equal.

75 + 75 + x = 180

x + 150 = 180

x + 150 - 150 = 180 - 150

x = 180 - 150

x = 30

2m is supplementary to the bottom right angle of the triangle.

that means they add together to make 180.

80 + 30 + (180 - 2m) = 180

110 - 2m = 180 - 180

110 - 2m = 0

110 = 2m

110/2 = m

55 = m

to save time I'm not going to write out the steps, its the same as the i.

2x + 3x + (180 - 120 ) = 180

5x + 60 = 180

5x = 180 - 60

5x = 120

x = 120/5

x = 24

imagine this as a triangle and a square.

sorry I'm not sure how to explain this -.-

but the equation would be

k + k + 90 = 180

2k + 90 = 180

2k = 180 - 90

2k = 90

k = 45

Im not sure how to solve this, I'm sorry :( hope you figure it out!

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2 years ago

At Western University the historical mean of scholarship examination scores for freshman applications is 900. A historical popul

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Answer:

a) Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b) The 95% confidence interval would be given by (910.05;959.95)

c) Since we confidence interval not ocntains the value of 900 we fail to reject the null hypothesis that the true mean is 900.

d) $z=\frac{935 -900}{\frac{180}{\sqrt{200}}}=2.750$

Since is a bilateral test the p value is given by:

$p_v =2*P(Z>2.750)=0.0059$

Step-by-step explanation:

a. State the hypotheses.

On this case we want to check the following system of hypothesis:

Null Hypothesis: $\mu =900$

Alternative hypothesis: $\mu \neq 900$

b. What is the 95% confidence interval estimate of the population mean examination score if a sample of 200 applications provided a sample mean x¯¯¯= 935?

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X=935$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma=180$ represent the population standard deviation

n=200 represent the sample size

The confidence interval for the mean is given by the following formula:

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

In order to calculate the mean and the sample deviation we can use the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$ (2)

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$ (3)

The mean calculated for this case is $\bar X=3278.222$

The sample deviation calculated $s=97.054$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$