Ask question

Ask question

All categories

1 year ago

6

Hi guys it me from earlier my device kicked me out what is an intiger?

2 answers:

Ira Lisetskai [31]1 year ago

8 0

Answer: an intiger is a whole number, positive or negative. It has no decimals.

Step-by-step explanation:

Darya [45]1 year ago

7 0

Answer: here it is!

Step-by-step explanation :

peter griffin is in fortnight thats the answer we all need

You might be interested in

What's 5 + 5 × 10 - 88 + 200 - 1

Mars2501 [29]

Answer:

=50x+116

Step-by-step explanation:

5+5x(10)−88+200−1

=5+50x+−88+200+−1

Combine Like Terms:

=5+50x+−88+200+−1

=(50x)+(5+−88+200+−1)

=50x+116

8 0

2 years ago

Read 2 more answers

Evgesh-ka [11]

The answer will end up being c

6 0

3 years ago

We have 6 face red cards (JQK x 2) and 16 black cards, so a total of 22 cards.The red face cards and the black cards numbered 2-

In-s [12.5K]

Answer:

I dont know

Step-by-step explanation:

only doing it for points

5 0

2 years ago

Plz help!!!!!!!!!!!!!!!!

Goshia [24]

Answer:

b) 95

Step-by-step explanation:

equation:

3 x 30 + 5

solve:

3 x 30 + 5 = 95

3 0

2 years ago

Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample

puteri [66]

Answer:

i $\to$ a

$n = 96040000$

i $\to$ b

$n_1 =24010000$

i $\to$ c

$n_2 =41602500$

ii $\to$ a

$E = 58.16$

ii $\to$ b

$291.84 < \mu < 408.16$ \

ii $\to$ c

There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

The sample size is $n = 8$

The sample mean is $\= x = \$ 350$

The sample standard deviation is $\$ 100$

Considering question i

i $\to$ a

At $E = 0.02$

given that the confidence level is 95% = 0.95

the level of significance would be $\alpha =1-0.95 = 0.05$

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

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

So the sample size is mathematically evaluated as

$n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=> $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=> $n = 96040000$

i $\to$ b

At $E_1 = 0.04$ and confidence level = 95% => $\alpha_1 = 0.05$ => $Z_{\frac{\alpha_1 }{2} } = 1.96$

$n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=> $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=> $n_1 =24010000$

i $\to$ c

At $E_2 = 0.04$ confidence level = 99% => $\alpha_2 = 0.01$

The critical value of $\frac{\alpha_2 }{2}$ from the normal distribution table is

$Z_{\frac{ \alpha_2 }{2} } = 2.58$

=> $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=> $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=> $n_2 =41602500$

Considering ii

Given that the level of significance is $\alpha = 0.10$

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

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

Generally the margin of error is mathematically represented as

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

substituting values

$E = 1.645 * \frac{100 }{\sqrt{8} }$

$E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

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

=> $350 - 58.16 < \mu < 350 + 58.16$

=> $291.84 < \mu < 408.16$

So the interpretation is that there is 90% confidence that the mean fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

8 0

2 years ago