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3 years ago

What are the greatest common factors of 48a^6 and 38a?

Mathematics

2 answers:

professor190 [17]3 years ago

5 0

The greatest common factor is 2a

3241004551 [841]3 years ago

3 0

Answer:

Step-by-step explanation:

38 is 19 multiplied by 2. 48 is 12 multiplied by 4. 6 multiplied by 8. 24 multiplied by 2. So the greatest common factor is 2a because there was no exponent on 38a

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You have $30 to buy muffins. However, at the shop you can buy 1 muffin for $4. (a) Write an inequality from the above informatio

sattari [20]

Answer:

Let m be the number of muffins.

a) 4m <= 30

b) m <= 7.5, so the maximum number of muffins to buy is 7.

c) $4 × 7 = $28, so $2 will be left over.

8 0

1 year ago

Choose the graph which represents the solution to the inequality:<br> 3x + 8 ≥ 11

Mashutka [201]

The solution of the inequality is x ≥ 1.

Solution:

Given inequality:

$3 x+8 \geq 11$

To find the solution of the inequality:

$3 x+8 \geq 11$

Subtract 8 from both sides.

$3 x+8-8 \geq 11-8$

$3 x \geq 3$

Divide by 3 on both sides.

$$\frac{3 x}{3} \geq \frac{3}{3}$

On simplifying this, we get

$x \geq 1$

The solution of the inequality is x ≥ 1.

The graph of the solution is attached below.

8 0

3 years ago

K.Brew sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet. R

Elan Coil [88]

Answer:

Step-by-step explanation:

The formula for determining the confidence interval for the difference of two population means is expressed as

Confidence interval = (x1 - x2) ± z√(s²/n1 + s2²/n2)

Where

x1 = mean sale amount for mail order sales = 82.70

x2 = mean sale amount for internet sales = 66.9

s1 = sample standard deviation for mail order sales = 16.25

s2 = sample standard deviation for internet sales = 20.25

n1 = number of mail order sales = 17

n2 = number of internet sales = 10

For a 99% confidence interval, we would determine the z score from the t distribution table because the number of samples are small

Degree of freedom =

(n1 - 1) + (n2 - 1) = (17 - 1) + (10 - 1) = 25

z = 2.787

Margin of error =

z√(s²/n1 + s2²/n2) = 2.787√(16.25²/17 + 20.25²/10) = 20.956190

4 0

3 years ago

The Information Technology Department at a large university wishes to estimate the proportion of students living in the dormitor

nadya68 [22]

Answer:

$n=\frac{0.5(1-0.5)}{(\frac{0.05}{1.96})^2}=384.16$

And rounded up we have that n=385

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . And the critical value would be given by: