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3 years ago

What is 25.6 divided by 0.5 =

Mathematics

2 answers:

mamaluj [8]3 years ago

6 0

51.2 it says on my calculator.

I hope this helps you alot. God Bless you.

N76 [4]3 years ago

4 0

It is 51.2 because yeah that’s is what I got in my calculator hope it helps (:

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suppose a normal distribution has a mean of 20 and a standard deviation of 4 a value of 26 is how many standard deviations away

marysya [2.9K]

Logically you can see that since 26 is half way between 24&28 it is 1/2 sd. See sketch

5 0

3 years ago

His composite shape is a rectangle with a semicircle attached on one end. The diameter of the semicircle is 6 feet.

Vera_Pavlovna [14]

We can solve by figuring the area of the rectangle then add it to the area of the semicircle.

Area of the rectangle: length · width = 6 · 10 = 60 ft²

Area of the semicircle: (πd²)/8 = (3.14 · 6²)/8 = (3.14 · 36)/8 = 14.13 ft²

Total area of the figure: 60 + 14.13 = 74.13 ft² ≈ 74 ft²

4 0

3 years ago

Read 2 more answers

A yellow ball, a red ball, a white ball and a

OleMash [197]

Answer:

1/4 chance (this is equal to 25% as a percentage or 0.25 as a decimal)

Step-by-step explanation:

There are 4 balls and only one can be first. As there is only 1 yellow ball, there is 1/4 chance that the yellow ball is picked first.

4 0

2 years ago

If y ¤ x = y2x for all positive integers, then (3 ¤ 4) ¤ 2 =

aivan3 [116]

96 is the answer to that

7 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

3 years ago