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3 years ago

Given Q is between J and K, JQ is five times the length of JK and JK=42 find the length of JQ

Mathematics

1 answer:

Alex73 [517]3 years ago

8 0

Answer:

210

Step-by-step explanation:

JQ is 5 times the length of JK

● JQ = 5JK

JK = 42

● JQ = 5 × 42

● JQ = 210

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4 Advantage, Inc., a tennis equipment manufacturer, has variable costs of $0.60 per unit of product. In August, the volume of pr

Amiraneli [1.4K]

Answer: the fixed costs per month is $14400

Step-by-step explanation:

Total cost = fixed cost + variable cost.

The tennis equipment manufacturer, has variable costs of $0.60 per unit of product. In August, the volume of production was 27,000 units. This means that the total variable cost from producing 27000 units is

27000 × 0.6 = 16200

If the total production costs incurred were $30,600, then the fixed costs per month would be

30600 = fixed cost + 16200

Fixed cost = 30600 - 16200

Fixed cost = $14400

6 0

3 years ago

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Solve. Enter your answer as a mixed number. The driving distance from Alex's house to the museum is 6 9 10 miles. What is the ro

givi [52]

Answer:

13.8 miles

Step-by-step explanation:

The driving distance from Alex's house to the museum is $6\frac{9}{10}$ miles

Since the distance from his house to the museum does not change, his trip back is also $6\frac{9}{10}$ miles.

The Round-Trip Distance is the distance to and from the museum.

To determine this, there ate two approaches.

Add the distances together
Multiply the distance to by 2.

Approach 1

$6\frac{9}{10}+6\frac{9}{10}\\\\=\frac{69}{10} + \frac{69}{10} \\=\frac{69+69}{10}\\=\frac{138}{10}\\=13\frac{4}{5}=13.8$

Approach 2

$6\frac{9}{10} X 2 \\=\frac{69}{10} X 2\\=\frac{69X2}{10}\\=\frac{138}{10}\\=13.8 miles$

The round trip distance is 13.8 miles.

5 0

3 years ago

Read 2 more answers

Frank joined a fitness club. He paid a monthly $50 registration fee, and dues are $12 per month.

brilliants [131]

Answer:

the answer is 50+12x. is this iready

3 0

2 years ago

a x − y = b a ⁢ x − y = b Which of the following equations are equivalent to the equation above? Select all that apply.

krek1111 [17]

They are all equivalent

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3 years ago

A researcher estimates the 90% CI for a sample with a mean of M = 8.9 and a standard error (sigma_M) of 2.1. What is the confide

Shalnov [3]

Answer:

The confidence interval at this level of confidence is between 5.4455 and 12.3545.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.05 = 0.95$ , so $z = 1.645$

Now, find M as such

$M = z*s$

In which s is the standard deviation of the sample, which is also called standard error. So

$M = 1.645*2.1 = 3.4545$

The lower end of the interval is the sample mean subtracted by M. So it is 8.9 - 3.4545 = 5.4455

The upper end of the interval is the sample mean added to M. So it is 8.9 + 3.4545 = 12.3545

The confidence interval at this level of confidence is between 5.4455 and 12.3545.

4 0

3 years ago