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4 years ago

Twice a number is twenty-eight

Mathematics

1 answer:

Alla [95]4 years ago

4 0

Let, the number = n
It would be: 2n = 28
n = 28/2
n = 14

In short, Your Answer would be 14

Hope this helps!

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For a given geometric sequence, the 9th term, a9, is equal to 43/256 and the 14th term, a14, is equal to 172. Find the value of

NeX [460]

Maybe solve by dividing. I think 5/6. Sorry math isn’t my strong suit. I’m good at reading.

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4 years ago

. A discount brokerage selected a random sample of 64 customers and reviewed the value of their accounts. The mean was $32,000 w

koban [17]

Answer:

The 90% confidence interval is $\$ \ 30313.9< \mu < \$ \ 33686.13$

Step-by-step explanation:

From the question we are told that

The sample size is n = 64

The sample mean is $\= x = \$ 32, 000$

The standard deviation is $\sigma= \$ 8, 200$

Given that the confidence interval is 90% then the level of significance is mathematically 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

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

Generally the margin of error is mathematically represented as

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

=> $E = 1.645 * \frac{ 8200 }{ \sqrt{64} }$

=> $E = 1686.13$

The 90% confidence interval is mathematically represented as

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

=> $32000 - 1689.13 < \mu < 32000 + 1689.13$

=> $\$ \ 30313.9< \mu < \$ \ 33686.13$

5 0

4 years ago

Male and female high school students reported how many hours they worked each week in summer jobs. The data is represented in th

kiruha [24]

The biggest key here is identifying outliers. Outliers can pull measures of center either up or down drastically.Notice how the males have a VERY low outlier. This will pull the mean down severely and create a much larger range, or spread.The females have a median of about 12 hours, but have an upper extreme of 19, which compareatively is very high. This creates a larger range as well.

7 0

3 years ago

Read 2 more answers

Cos^2(2theta)+ 2cos(2theta)+1=0

Tema [17]

Answer:

θ = π n + π/2 for n element Z

Step-by-step explanation:

Solve for θ:

1 + 2 cos(2 θ) + cos^2(2 θ) = 0

Write the left hand side as a square:

(cos(2 θ) + 1)^2 = 0

Take the square root of both sides:

cos(2 θ) + 1 = 0

Subtract 1 from both sides:

cos(2 θ) = -1

Take the inverse cosine of both sides:

2 θ = 2 π n + π for n element Z

Divide both sides by 2:

Answer: θ = π n + π/2 for n element Z

4 0

3 years ago

An SAT prep course claims to improve the test score of students. The table below shows the scores for seven students the first t

poizon [28]

Answer:

Since the calculated value of t = -0.215 falls in the critical region so we accept Ha that SAT prep course improves the students' verbal SAT scores and reject the null hypothesis at significance level 0.05.

e. These results support the claim that the SAT prep course improves the students' verbal SAT scores.

Step-by-step explanation:

Student 1 2 3 4 5 6 7

Score on first SAT 500 380 560 430 450 360 560

Score on second SAT 540 470 580 450 480 400 600

Difference d -40 -90 -20 -20 -30 -40 -40 ∑ -280

d² 1600 8100 400 400 900 1600 1600 ∑14600

a. Let the hypotheses be

H0: ud= 0 against the claim Ha: ud ≠0

The degrees of freedom = n-1= 7-1= 6

The significance level is 0.05

The test statistic is

t= d`/sd/√n

The critical region is ║t║≤ t (0.025,6) = ±2.447

d`= ∑di/n= -280/7= -4

Sd²= ∑(di-d`)²/n-1 = 1/n-1 [∑di²- (∑di)²n]

= 1/6[14600-(-4)²/7] = [14600-2.2857/6]= 2432.952

b. Sd= 49.3249= 49.325

Therefore

c. t= d`/ sd/√n

t= -4/ 49.325/√7

t= -4/18.6435 = -0.2145= -0.215

d. Since the calculated value of t = -0.215 falls in the critical region so we accept Ha that SAT prep course improves the students' verbal SAT scores and reject the null hypothesis at significance level 0.05

e. These results support the claim that the SAT prep course improves the students' verbal SAT scores.

6 0

3 years ago