Ask question

Ask question

All categories

3 years ago

5

Janet can make 4/5 of a necklace in 20 minutes at this straight how many necklaces to the nearest 10th of a necklace can Janet m

ake in one hour.
Please help

1 answer:

ziro4ka [17]3 years ago

4 0

12/15 i hope this helps , 20 goes into one hour 3 times so i multiplied 4/5 x 3

You might be interested in

Stuck in it from a week!!<br> Help please.

nadya68 [22]

The answer is 49 because x = 1/7 and 1/7^-2 =49

5 0

4 years ago

Jese hangs 3 t shirt on a line. any t shirt next to each other share a peg. he uses 4 pegs. how may peg jess use to hang 6 t shi

Ivenika [448]

He will use 7 pegs in total to hang 6 t shirts in line

7 0

4 years ago

within a kiambu county,students were randomly assigned to one of two mathematics teachers.Mrs.Elite and Mrs. Bright. After the a

DerKrebs [107]

Using the t-distribution, as we have the standard deviation for the samples, it is found that since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, we test if they are equally effective teachers, that is, the subtraction of their means is 0, hence:

$H_0: \mu_1 - \mu_2 = 0$

At the alternative hypothesis, we test if they are not equally effective teachers, that is, the subtraction of their means is not 0, hence:

$H_1: \mu_1 - \mu_2 \neq 0$

<h3>What is the distribution of the differences?</h3>

For Mrs. Elite, we have that:

$\mu_1 = 78, \sigma_1 = 10, n_1 = 30, s_1 = \frac{10}{\sqrt{30}} = 1.82574$

For Mrs. Bright, we have that:

$\mu_2 = 85, \sigma_2 = 15, n_2 = 25, s_2 = \frac{15}{\sqrt{25}} = 3$

For the distribution of differences, we have that:

$\overline{x} = \mu_1 - \mu_2 = 78 - 85 = -7$

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{1.82574^2 + 3^2} = 3.5119$

<h3>What is the test statistic?</h3>

The test statistic is given by:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{-7 - 0}{3.5119}$

$t = -1.99$

Considering a<em> two-tailed test</em>, as we are testing if the mean is different of a value, with 30 + 25 - 2 = <em>53 df and a significance level of 0.1</em>, the critical value is of $|t^{\ast}| = 1.6741$ .

Since the absolute value of the test statistic is greater than the critical value, we reject the claim that Mrs.Elite and Mrs. Bright are equally effective teachers.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

3 0

3 years ago

A)distributive property of multiplication over subtraction

alina1380 [7]

Answer: sup brendalys it’s me Timotheus

Step-by-step explanation:

The answer is c I think

5 0

3 years ago

What is the value of the expression shown below

kaheart [24]

value of the expression $\frac{1.6(10^{5})}{0.2(10^{2})}$ or $1.6(10^{5})$ over $0.2(10^{2})$ is $8000$ ( $8 \times 10 ^3$ ).

<u>Step-by-step explanation:</u>

Here we have , to find the value of the expression shown below 1.6 x 10 to the 5th power over 0.2 x 10 to the 2nd power i.e. $1.6(10^{5})$ over $0.2(10^{2})$ :

⇒ $1.6(10^{5})$ over $0.2(10^{2})$

⇒ $\frac{1.6(10^{5})}{0.2(10^{2})}$

multiply and divide by 10 to remove decimals, so we get

⇒ $\frac{16(10^{5})}{2(10^{2})}$

⇒ $\frac{8(2)(10^{5})}{2(10^{2})}$

⇒ $8\frac{(10^{5})}{(10^{2})}$

⇒ $8(10^{(5-2)})$

⇒ $8(10^{3})$

⇒ $8(1000)$

⇒ $8000$

Therefore, value of the expression $\frac{1.6(10^{5})}{0.2(10^{2})}$ or $1.6(10^{5})$ over $0.2(10^{2})$ is $8000$ ( $8 \times 10 ^3$ ).

7 0

4 years ago

Read 2 more answers