Ask question

Ask question

All categories

2 years ago

15

Rename the number 4 thousand 7 hundreds

2 answers:

Paraphin [41]2 years ago

8 0

4,700 hope u get it correct

Brrunno [24]2 years ago

4 0

The number would be written numerically like so:

4,700

You might be interested in

(-59)-(-18)/(+3)+(-2)(+4)+-27-(-3)

mojhsa [17]

-41/25 I believe, I'm not sure why you put some addition signs in the inside of parentheses

8 0

2 years ago

Read 2 more answers

The following results come from two independent random samples taken of two populations.

photoshop1234 [79]

Answer:

$(a)\ \bar x_1 - \bar x_2 = 2.0$

$(b)\ CI =(1.0542,2.9458)$

$(c)\ CI = (0.8730,2.1270)$

Step-by-step explanation:

Given

$n_1 = 60$ $n_2 = 35$

$\bar x_1 = 13.6$ $\bar x_2 = 11.6$

$\sigma_1 = 2.1$ $\sigma_2 = 3$

Solving (a): Point estimate of difference of mean

This is calculated as: $\bar x_1 - \bar x_2$

$\bar x_1 - \bar x_2 = 13.6 - 11.6$

$\bar x_1 - \bar x_2 = 2.0$

Solving (b): 90% confidence interval

We have:

$c = 90\%$

$c = 0.90$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.90$

$\alpha = 0.10$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.10/2}$

$z_{\alpha/2} = z_{0.05}$

The z score is:

$z_{\alpha/2} = z_{0.05} =1.645$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.645 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.645 * \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.645 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.645 * \sqrt{0.3306}$

$2.0 \± 0.9458$

Split

$(2.0 - 0.9458) \to (2.0 + 0.9458)$

$(1.0542) \to (2.9458)$

Hence, the 90% confidence interval is:

$CI =(1.0542,2.9458)$

Solving (c): 95% confidence interval

We have:

$c = 95\%$

$c = 0.95$

Confidence level is: $1 - \alpha$

$1 - \alpha = c$

$1 - \alpha = 0.95$

$\alpha = 0.05$

Calculate $z_{\alpha/2}$

$z_{\alpha/2} = z_{0.05/2}$

$z_{\alpha/2} = z_{0.025}$

The z score is:

$z_{\alpha/2} = z_{0.025} =1.96$

The endpoints of the confidence level is:

$(\bar x_1 - \bar x_2) \± z_{\alpha/2} * \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}$

$2.0 \± 1.96 * \sqrt{\frac{2.1^2}{60}+\frac{3^2}{35}}$

$2.0 \± 1.96* \sqrt{\frac{4.41}{60}+\frac{9}{35}}$

$2.0 \± 1.96 * \sqrt{0.0735+0.2571}$

$2.0 \± 1.96* \sqrt{0.3306}$

$2.0 \± 1.1270$

Split

$(2.0 - 1.1270) \to (2.0 + 1.1270)$

$(0.8730) \to (2.1270)$

Hence, the 95% confidence interval is:

$CI = (0.8730,2.1270)$

8 0

3 years ago

3(x+6)-2=28 to 3(x+6)=30 associative commutative or distributive

tekilochka [14]

Answer:

associative

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Gloria received a 4 percent raise and is now making $24,960 a year, what was her salary before the raise?

Svetradugi [14.3K]

She gets a 4% raise so her new pay is 100% + 4% of her previous pay.

104% = 1.04 as a decimal.

Divide her new pay by 1.04:

24,960 / 1.04 = 24,000

Her previous pay was $24,000

5 0

2 years ago

Write a word problem for which you would use the equation 2/5x + 6 = 56

velikii [3]

Answer:

Isha is at an arcade. There is a base fee of $6 to enter the game room. Every game costs $0.4 or 40 cents to play. If Isha ends up spending $56 at the arcade, how many games did she play?

An explanation:

The base fee of $6 represents the constant 6 in the equation. Every game costs $0.4, which equals 2/5.

Hope this helps! If you don't like this one I might be able to think of another one :)

7 0

3 years ago