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White raven [17]

3 years ago

12

Help me please !!!!!

1 answer:

s2008m [1.1K]3 years ago

4 0

I think it’s either C or A but more than likely C

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Solve This Equation 5x - 5 = 50

Black_prince [1.1K]

X=11
All you have to do is add 5 to both sides and then divide by 5 on both sides

7 0

3 years ago

Read 2 more answers

Find the mean, median, and mode(s) of the data. Choose the measure that best represents the data. Explain your reasoning. Find t

Kazeer [188]

Answer:

Mean: 17

Median: 16

Mode: 10

Step-by-step explanation:

8, 10, 10, 11, 16, 17, 19, 21, 41

Median: (the middle number in the number list/data set) 16

Mode: 10

Mean: 8 + 10 + 10 + 11 + 16 + 17 + 19 + 21 + 41 = 153/9 = 17

3 0

3 years ago

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

Vikentia [17]

Answer:

a

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

b

$15.505 < \mu < 21.495$

c

$14.93 < \mu < 22.069$

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

Is 2 a solution of the inequality 4 + a ≤ 6? *

zaharov [31]

Answer:

The answer is "Yes, 6 = 6"

Step-by-step explanation:

If you substitute 2 for A, you would get 4 + 2 is less than or equal to 6, which is true because 6 = 6.

7 0

2 years ago

Please answer correctly if you know. (No links please).

eimsori [14]

Answer:

The Answer is $x^{\frac{7}{3\\} }$

Step-by-step explanation:

Let me know if i got this right...? :)

7 0

2 years ago