What is the distributive property of 1,107×2

Please explain me and tell me the right answer.

Shkiper50 [21]

Basically, it asks you to find the piece of data that is 1/10th of the overall shoe sales on a given day. As the number of shoes sold per day is 400, the piece of data you are looking for is 40. Meaning the answer is 'High heels sold on Friday'

4 0

3 years ago

A manufacturing company with 450 employees begins a new product line and must increase their number of employees by 18%. How man

____ [38]

The total employees the company now have is 531.

Solution:

Beginning employees in a company = 450

Percentage of employees increased = 18%

Number of increased employees = 18% of 450

$$=\frac{18}{100}\times 450$

= 81

Total employees = Beginning employees + Increased employees

= 450 + 81

= 531

The total employees the company now have is 531.

3 0

3 years ago

Y=−2x+4<br> Complete the missing value in the solution to the equation.<br><br> (____,-2)

mestny [16]

(3, - 2)

given y = - 2, then

- 2x + 4 = - 2 ( solve for x )

subtract 4 from both sides

- 2x = - 6 ( divide both sides by - 2 )

x = $\frac{-6}{-2}$ = 3

3 0

3 years ago

Read 2 more answers

The following data comparing wait times at two rides at Disney are listed below: Position Pirates Splash Mountain Sample Size 32

myrzilka [38]

Answer:

a) $(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

b) $t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

Step-by-step explanation:

Data given and notation

$\bar X_{A}=14.68$ represent the mean for Pirates

$\bar X_{B}=18.77$ represent the mean for Splash Mountain

$s_{A}=11.87$ represent the sample standard deviation for the sample Pirates

$s_{B}=16.79$ represent the sample standard deviation for the sample Slpash Mountain

$n_{A}=32$ sample size selected for Pirates

$n_{B}=30$ sample size selected for Splash Mountain

$\alpha=0.02$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

Part a

The confidence interval would be given by:

$(\bar X_A -\bar X_B) \pm t_{\alpha/2} \sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}$

The degrees of freedom are given by:

$df = n_A +n_B -2 = 32+30-2 = 60$

Since we want 98% of confidence the significance level is $\alpha =1-0.98 =0.02$ and $\alpha/2 =0.01$ , we can find in the t distribution with df =60 a critical value that accumulates 0.01 of the area on each tail and we got:

$t_{\alpha/2}= 2.39$

And replacing we got for the confidence interval:

$(14.68 -18.77) - 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =-12.968$

$(14.68 -18.77) + 2.39 \sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}} =4.788$

Part b

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the means are equal, the system of hypothesis would be:

Null hypothesis: $\mu_{A} = \mu_{B}$

Alternative hypothesis: $\mu_{A} \neq \mu_{B}$

the statistic is given by:

$t=\frac{\bar X_{A}-\bar X_{B}}{\sqrt{\frac{s^2_{A}}{n_{A}}+\frac{s^2_{B}}{n_{B}}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{14.68-18.77}{\sqrt{\frac{11.87^2}{32}+\frac{16.79^2}{30}}}}=-1.10$

6 0

3 years ago

Simplify (x⁴)³ ??????

Morgarella [4.7K]

Rule:

(xᵃ)ᵇ = x⁽ᵃ * ᵇ⁾

(x⁴)³ = x⁽⁴ * ³⁾ = x¹²

3 0

3 years ago