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3 years ago

Complete this equations,( show your work).

Mathematics

1 answer:

timurjin [86]3 years ago

3 0

Answer:

21 is: $\frac{4}{9}$ x + $\frac{31}{72}$

22 is: 4 $\frac{39}{44}$

23 is: $\frac{1}{4}$ x + $\frac{31}{72}$

24 is: $\frac{-2}{11}$ x + $\frac{7}{11}$

Step-by-step explanation: Hope this helps and good luck mark brainliest if u want :)

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(PLEASE HELP) <br> Find the area of the parallelogram.

allochka39001 [22]

Answer:

Step-by-step explanation:

A = bh

B = 7

H=10

7x10 =70m

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2 years ago

I need help on this one

shusha [124]

Answer:

= 1/26

Step-by-step explanation:

Use Division Distributive Property:

(x/y)^a = x^a/y^a

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3 years ago

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2 years ago

A machine can manufacture 39 units per hour. Assuming 8 hours per day and 5 days per week, the manufacturing time per unit is mi

Crank

Answer:

The manufacturing time is 1.5385 minutes per unit.

Step-by-step explanation:

∵ The units manufactured by machine in 1 hour = 39,

We know that,

1 hour = 60 minutes,

∴ The units manufactured by machine in 60 minutes = 39,

That is,

39 units = 60 minutes,

⇒ 1 unit = $\frac{60}{39}$ =1.53846153846 ≈ 1.5385 minutes,

Hence, the manufacturing time is 1.5385 minutes per unit.

8 0

4 years ago

A simple random sample of 20 days in which Parsnip ate seeds was selected, and the mean amount of time it took him to eat the se

bearhunter [10]

Answer:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

Step-by-step explanation:

We know the following info from the problem

$\bar X_1 = 14.5$ sample mean for the group 1

$s_1 = 3.98$ the standard deviation for the group 1

$n_1= 20$ the sample size for group 1

$\bar X_2 = 13.9$ sample mean for the group 2

$s_1 = 4.03$ the standard deviation for the group 2

$n_2= 17$ the sample size for group 2

We have all the conditions satisifed since we have random samples.

We want to construct a confidence interval for the true difference of means and the correct formula for this case is:

$(\bar X_1 -\bar X_2) \pm t_{\alpha/2}\sqrt{\frac{s^2_1}{n_1} +\frac{s^2_2}{n_2}}$

The degrees of freedom are given :

$df = n_1 +n_2- 2 = 20+17-2=35$

The confidence level is 0.9 or 90% and the significance level is $\alpha=1-0.9=0.1$ and $\alpha/2 = 0.05$ and the critical value for this case is:

$t_{\alpha/2} = 1.69$

And replacing the info given we got:

$(14.5-13.9) -1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = -1.63$

$(14.5-13.9) +1.69 \sqrt{\frac{3.98^2}{20}+ \frac{4.03^2}{17}} = 2.83$

And the confidence interval for this case $-1.63 \leq \mu_1 -\mu_2 \leq 2.83$

5 0

3 years ago