50 people sponsor paula and she raises a total of £180 for charity. what is the mean average amount sponsored?

Graph the solution on a number line -6x+3>-39

SCORPION-xisa [38]

Step-by-step explanation:

$-6x+3>-39\qquad\text{subtract 3 from both sides}\\\\-6x+3-3>-39-3\\\\-6x>-42\qquad\text{change the all signs}\\\\6x$

$-\text{op}\text{en circle}-\circ\\\leq,\ \geq-\text{closed circle}-\bullet\\\\,\ \geq-\text{draw the line to the right}$

5 0

3 years ago

A manufacturer of paper coffee cups would like to estimate the proportion of cups that are defective (tears, broken seems, etc.)

Maksim231197 [3]

Answer:

a

The 95% confidence interval is $0.0503 < p < 0.1297$

b

The sample proportion is $\r p = 0.09$

c

The critical value is $Z_{\frac{\alpha }{2} } = 1.96$

d

The standard error is $SE =0.020$

Step-by-step explanation:

From the question we are told that

The sample size is n = 200

The number of defective is k = 18

The null hypothesis is $H_o : p = 0.08$

The alternative hypothesis is $H_a : p > 0.08$

Generally the sample proportion is mathematically evaluated as

$\r p = \frac{18}{200}$

$\r p = 0.09$

Given that the confidence level is 95% then the level of significance is mathematically 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

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

Generally the standard of error is mathematically represented as

$SE = \sqrt{\frac{\r p (1 - \r p)}{n} }$

substituting values

$SE = \sqrt{\frac{0.09 (1 - 0.09)}{200} }$

$SE =0.020$

The margin of error is

$E = Z_{\frac{ \alpha }{2} } * SE$

=> $E = 1.96 * 0.020$

=> $E = 0.0397$

The 95% confidence interval is mathematically represented as

$\r p - E < \mu < p < \r p + E$

=> $0.09 - 0.0397 < \mu < p < 0.09 + 0.0397$

=> $0.0503 < p < 0.1297$

7 0

3 years ago

What is the best estimate for the length of a bulletin board in meters

Reptile [31]

Bulletin boards can be in all shapes and sizes. There is no correct or incorrect answer to this problem.

6 0

3 years ago

Read 2 more answers

A manufacturer of new light bulbs claims the average lifetime of its long-life bulb is more than 4000 hours. To test this claim,

natka813 [3]

Answer: C. 12.5

Step-by-step explanation:

Given : A manufacturer of new light bulbs claims the average lifetime of its long-life bulb is more than 4000 hours.

Population mean : $\mu=4000$

Sample size : n= 100

Sample mean : $\overline[x}=4500$

Standard deviation: $s=400$

The value of test-statistic is given by :-

$z=\dfrac{\overline{x}-\mu}{\dfrac{s}{\sqrt{n}}}\\\\\Rightarrow\ z=\dfrac{4500-4000}{\dfrac{400}{\sqrt{100}}}\\\\\Rightarrow\ z= 12.5$

Hence, the value of the test statistic for this problem is 12.5.

5 0

3 years ago

I don't know the answer marked stylish random

Kamila [148]

Answer:

A

Step-by-step explanation: i am in math 8

lol

8 0

2 years ago