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2 years ago

Please help me i know its a hard question but please help i need the answer asap

Mathematics

1 answer:

Temka [501]2 years ago

7 0

Answer:

1. Angle ABC = Angle DEF

2. Angle BCA = Angle EFD

3. Segment AC = Segment DF

4. Segment BA = Segment ED

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Idk wat to do.....<br>........

oksano4ka [1.4K]

Answer:

Mean : 95

Median : 85

Mode : 90

Part B : Impossible

Step-by-step explanation:

We can make an equation to find the mean using the first 5 history test scores.

$85=\frac{80+75+90+75+95+n}{6} \\ \\ 85=\frac{415+n}{6} \\ \\ 510=415+n \\ \\ n=95$

So a 95 would be needed to have a mean of 85.

Next, the median.

First, we sort the first 5 history scores from least to greatest.

We get 75, 75, 80, 90, 95.

Since, 80 is the middle value, it will be used in the calculation of the median.

We can make an equation with this.

$82.5=\frac{80+n}{2} \\ \\ 165=80+n \\ \\ n=85$

So a score a 85 would be needed to have a median of 82.5

Thirdly, the mode.

Since 90 is already in the set once, we can just have Maliah score another 90 to make 90 the mode (with the exception of 75 of course).

Finally, Part B.

We can use the equation we had for the first mean calculation but change 85 to 90.

$90=\frac{80+75+90+75+95+n}{6} \\ \\ 90=\frac{415+n}{6} \\ \\ 540=415+n \\ \\ n=125$

So Maliah would need a score of 125 to make her mean score 90, but since the range is only from 0-100, it is impossible.

8 0

3 years ago

I WILL GIVE BRAINLEST!!! 20 POINTS!!!

Alenkasestr [34]

It the answer would be 51.41

3 0

2 years ago

Read 2 more answers

Let the following sample of 8 observations be drawn from a normal population with unknown mean and standard deviation:

Vikentia [17]

Answer:

$\= x = 18.5$ , $\sigma = 5.15$

$15.505 < \mu < 21.495$

$14.93 < \mu < 22.069$

Step-by-step explanation:

From the question we are are told that

The sample data is 21, 14, 13, 24, 17, 22, 25, 12

The sample size is n = 8

Generally the ample mean is evaluated as

$\= x = \frac{\sum x }{n}$

$\= x = \frac{ 21 + 14 + 13 + 24 + 17 + 22+ 25 + 12 }{8}$

$\= x = 18.5$

Generally the standard deviation is mathematically evaluated as

$\sigma = \sqrt{\frac{\sum (x- \=x )^2}{n}}$

$\sigma = \sqrt{\frac{\sum ((21 - 18.5)^2 + (14-18.5)^2+ (13-18.5)^2+ (24-18.5)^2+ (17-18.5)^2+ (22-18.5)^2+ (25-18.5)^2+ (12 -18.5)^2 )}{8}}$

$\sigma = 5.15$

considering part b

Given that the confidence level is 90% then the significance level is evaluated as

$\alpha = 100-90$

$\alpha = 10\%$

$\alpha = 0.10$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.645$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.645 * \frac{5.15 }{\sqrt{8} }$

=> $E = 2.995$

The 90% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 2.995 < \mu < 18.5 + 2.995$

$15.505 < \mu < 21.495$

considering part c

Given that the confidence level is 95% then the significance level is evaluated as

$\alpha = 100-95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{ \alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

=> $E =1.96 * \frac{5.15 }{\sqrt{8} }$

=> $E = 3.569$

The 95% confidence interval is evaluated as

$\= x - E < \mu < \= x + E$

substituting values

$18.5 - 3.569 < \mu < 18.5 + 3.569$

$14.93 < \mu < 22.069$

8 0

3 years ago

It cost $2.75 to rent bowling shoes each game cost $6.00 you have $14.75 how many games can you bowl

densk [106]

14.75 - 2.75 = 12.00

12.00 / 6.00 = 2

you can play 2 games

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3 years ago

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HELP QUICK!! 20 POINTS AND BRAINLIEST FOR CORRECT ANSWER!! In circle O, what is the m∠CAT?

il63 [147K]

Answer:D

Step-by-step explanation:

Your welcome

6 0

3 years ago

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