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posledela
2 years ago
14

Help me with this pls

Mathematics
2 answers:
zzz [600]2 years ago
6 0

Answer:

The answer should be D) 60 = (x + 45)

Step-by-step explanation:

bazaltina [42]2 years ago
3 0

Answer:

D. 60=(x+45)

Step-by-step explanation:

They are the same degree so they should equal eachother

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Please answer ASAP will mark brainilest ​
Otrada [13]

Answer:

1

Step-by-step explanation:

Change of Y is 0 and change of x is any number.

0/any number = 0

6 0
3 years ago
Find the area of the lateral faces of the right triangular prism with altitude of 5 cm and base edges of length 3 cm, 4 cm, and
Marysya12 [62]
Check the picture below.

the base of it is one of the triangles, and thus, the sides are just the 3 rectangles.

notice all you have is a 3x5 rectangle, a 5x5 one and a 4x5 rectangle.

get their area, sum them up, and that's the lateral, sides, area of the triangular prism.

(3 · 5)  +  (5 · 5)  +  (4 · 5)

6 0
2 years ago
Assume that population mean is to be estimated from the sample described. Use the sample results to approximate the margin of er
Vikentia [17]

Answer:

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

Step-by-step explanation:

Previous concepts

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 =69 represent the sample mean for the sample  

\mu population mean

s=4.0 represent the sample standard deviation

n=64 represent the sample size  

Assuming the X follows a normal distribution

X \sim N(\mu, \sigma)

The sample mean \bar X have the following distribution:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

The margin of error for the sample mean is given by this formula:

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

Since we have the sample standard deviation we can estimate the margin of error like this:

ME=t_{\alpha/2}\frac{s}{\sqrt{n}}

The next step would be find the value of \z_{\alpha/2}, \alpha=1-0.95=0.05 and \alpha/2=0.025

We can find the degrees of freedom like this:

df=n-1=64-1=63

And we can find the critical value with the following code in excel for example: "=T.INV(0.025,63)" and we got:

t_{\alpha/2}=\pm 1.998

And the margin of error would be :

ME=1.998 \frac{4}{\sqrt{64}}=0.999 \approx 1

3 0
3 years ago
What is the area of the rhombus shown below?
zzz [600]

Answer:

C

Step-by-step explanation:

i took test

5 0
3 years ago
Blake and James are planning to go on a trip. Blake already has some money saved. She earns $96 more by working overtime. James
rosijanka [135]

Blake saved \bold{\$ 433.80}

Option - A

<u>SOLUTION: </u>

Given, Blake and James are planning to go on a trip.  

Blake already has some money saved.  

Let the amount saved by Blake be \$ n.

She earns \$96 more by working overtime.  

Then, total amount by Blake =\$(n+96)

James has \$ 53 more than 4 times the amount of money that Blake has saved.  

Then, total amount by James will be \$ 53+4 \times \$ n=\$(4 n+53)

If they have \$2,318 in all, we have to find how much money had Blake originally saved

Now, total amount = \$2318

\begin{array}{l}{\rightarrow \text { Amount by Blake }+\text { amount by James }=2318} \\\\ {\rightarrow \mathrm{n}+96+4 \mathrm{n}+53=2318 \rightarrow 5 \mathrm{n}+149=2318 \rightarrow 5 \mathrm{n}=2318-149 \rightarrow 5 \mathrm{n}=2169 \rightarrow \mathrm{n}=433.8}\end{array}

7 0
3 years ago
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