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3 years ago

What is the least common denominator of 8 and 12

Mathematics

1 answer:

lisabon 2012 [21]3 years ago

6 0

Answer:

The least common denominator of 8and 12 is 24

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Write a linear function rule for the data in the table.<br><br> x 0 1 2 3 4<br> y 3 1 –1 –3 –5

Dima020 [189]

Answer:

f(x) = –2x + 3

Step-by-step explanation:

7 0

3 years ago

If Rib-eye steaks are on sale for $8.52 per pound. Ms. Markum bought a 0.9-pound steak. How much did the steak cost?

Pepsi [2]

Answer: The steak costs $ 7.67.

Step-by-step explanation:

Given: Cost per pound of Rib-eye steaks = $8.52

Ms. Markum bought a 0.9-pound steak.

To find : cost of 0.9 pound steak =

Then cost of 0.9 pound steak = 0.9 x (Cost per pound of Rib-eye steaks )

= $ (0.9 x 8.52)

= $7.668 ≈ $ 7.67

Hence, the steak costs $ 7.67.

7 0

3 years ago

29 is 6more than k ( write answer and equation)

almond37 [142]

K is 23.
As for the equation, it is
29=k+6
To solve, subtract 6 from both sides.

6 0

3 years ago

Read 2 more answers

In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol

Julli [10]

Answer:

$0.071,1.928$

Step-by-step explanation:

Downtown Store North Mall Store

Sample size n 25 20

Sample mean $\bar{x}$ $9 $8

Sample standard deviation s $2 $1

$n_1=25\\n_2=20$

$\bar{x_1}=9\\ \bar{x_2}=8$

$s_1=2\\s_2=1$

$x_1-x_2=9-8=1$

Standard error of difference of means = $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Standard error of difference of means = $\sqrt{\frac{2^2}{25}+\frac{1^2}{20}}$

Standard error of difference of means = $0.458$

Degree of freedom = $\frac{\sqrt{(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1}+\frac{(\frac{s_2^2}{n_2})^2}{n_2-1}}$

Degree of freedom = $\frac{\sqrt{(\frac{2^2}{25}+\frac{1^2}{20}})^2}{\frac{(\frac{2^2}{25})^2}{25-1}+\frac{(\frac{1^2}{20})^2}{20-1}}$

Degree of freedom =36

So, z value at 95% confidence interval and 36 degree of freedom = 2.0280

Confidence interval = $(x_1-x_2)-z \times SE(x_1-x_2),(x_1-x_2)+z \times SE(x_1-x_2)$

Confidence interval = $1-(2.0280)\times 0.458,1+(2.0280)\times 0.458$

Confidence interval = $0.071,1.928$

Hence Option A is true

Confidence interval is $0.071,1.928$

4 0

3 years ago

Hello! May someone please help on this :)<br> I would probably give brainliest.

ivanzaharov [21]

Answer:

theres 16 doughnuts

Step-by-step explanation:

6 0

3 years ago