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Tcecarenko [31]
3 years ago
14

How would I solve Part B?

Mathematics
1 answer:
alekssr [168]3 years ago
7 0

Answer:

D (-4, -9)

Step-by-step explanation:

midpoint formula:

(x₁ + x₂) / 2 , (y₁ + y₂) / 2

plug in C (2,7)

(2 + x₂) / 2 , (7 + y₂ ) / 2 = (-1, -1)

(2 + x₂) / 2 = -1

multiply each side by 2

2 + x₂ = -2

subtract 2 from both sides

x₂ = -4

(7 + y₂) / 2 = -1

multiply each side by 2

7 + y₂ = -2

subtract 7 from both sides

y₂ = -9

D (-4, -9)

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Will mark brainliest <br> Number 1 and 3
Keith_Richards [23]

Answer: Q1=20 Q2=39

Step-by-step explanation:

For question one all you have to do is multiply 20x35x8 giving you 5600. For question two just multiply 39x7x8 giving you 2184. Beat of luck.

8 0
2 years ago
An SRS of 350 high school seniors gained an average of x¯=21 points in their second attempt at the SAT Mathematics exam. Assume
Lynna [10]

Answer:

a) 21-2.58\frac{52}{\sqrt{350}}=13.83  

21+2.58\frac{52}{\sqrt{350}}=28.17  

So on this case the 99% confidence interval would be given by (13.83;28.17)  

b) ME=2.58\frac{52}{\sqrt{350}}=7.17

c) ME=2.58\frac{52}{\sqrt{100}}=13.42

d) D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

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".  

Part a

\bar X=21 represent the sample mean  

\mu population mean (variable of interest)  

\sigma=52 represent the population standard deviation  

n=350 represent the sample size  

99% confidence interval  

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

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

Since the Confidence is 0.99 or 99%, the value of \alpha=0.01 and \alpha/2 =0.005, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.005,0,1)".And we see that z_{\alpha/2}=2.58  

Now we have everything in order to replace into formula (1):  

21-2.58\frac{52}{\sqrt{350}}=13.83  

21+2.58\frac{52}{\sqrt{350}}=28.17  

So on this case the 99% confidence interval would be given by (13.83;28.17)  

Part b

The margin of error is given by:

ME=2.58\frac{52}{\sqrt{350}}=7.17

Part c

The margin of error is given by:

ME=2.58\frac{52}{\sqrt{100}}=13.42

Part d

As we can see when we reduce the sample size we increase the margin of error so the best option for this case is:

D. Decreasing the sample size increases the margin of error, provided the confidence level and population standard deviation remain the same.

6 0
2 years ago
How much will the guitar be worth in 10 years?
Contact [7]

Answer:

$44487

Step-by-step explanation:

If you plug in 10 for t, you are left with the following equation:

V=12000(1.14)^{10}=12000\cdot 3.7072\approx 44487. Hope this helps!

8 0
2 years ago
Read 2 more answers
What is the value of x?
vfiekz [6]

Answer:

80°

Step-by-step explanation:

angles in a triangle add up to 180°

70° + 30° = 100°

180° - 100° = 80°

7 0
3 years ago
Read 2 more answers
Please help me!!!! You only have to answer number 1!!!
mrs_skeptik [129]

Given : A social media website has currently = 1,000 members.

           In each month number of members gets triple.

To find: Equation that represents number of members m belongs to site.

Solution: Because we are given, each month the number of members gets triple, that means each month we can multiply by 3 to the number of members of previous month.

For first month, we would multiply 1,000 by 3.

For second month, we would multiply 1,000 by 3 *3.

For second month, we would multiply 1,000 by 3*3*3.

Each month one factor of 3 is being increasing.

We could write those 3's as exponents of 3.

Like 1000*3 = 1000*3^1

1000 * 3^2.

1000 * 3^3.

As many months as many 3's are there.

So, the exponent of 3 is represnts the number of months.

Let us take a variable for number of month by n.

So, we could setup an equation for number of members m belongs to site on a particular month n as

m= 1000 (3)^n

This is our final equation.

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