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Brilliant_brown [7]
3 years ago
15

Lacey pays $15 for each hour of golf lessons and $10 for equipment rental. If the lesson is more than 3 hours long, she only pay

s $5 for equipment rental. Is the total cost of Lacey’s lesson a function of the number of hours scheduled?
Mathematics
1 answer:
love history [14]3 years ago
5 0

Answer:

yes because 15*3 if she pays 15 for 3 hours

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(08.03)
Alik [6]

Answer:

Option D, (x + 2)(x - 6)

Step-by-step explanation:

<u>Step 1:  Factor </u>

x^2 - 4x - 12

x^2 - 6x + 2x - 12

(x - 6)(x + 2)

(x + 2)(x - 6)

Answer:  Option D, (x + 2)(x - 6)

8 0
3 years ago
Simplify the expression below using the distributive property
Rus_ich [418]

Answer:

-84-12i

Step-by-step explanation:

The one you have is the answer :)

3 0
3 years ago
Find the product of 2/5 with the quotient of 7/3 and 4/5
marissa [1.9K]

Answer:

7/6

Step-by-step explanation:

7/3÷4/5

7/3×5/4

35/12

Now the multiplication

2/5×35/12

70/60

7/6

6 0
3 years ago
An automobile manufacturer has given its van a 59.5 miles/gallon (MPG) rating. An independent testing firm has been contracted t
LekaFEV [45]

Answer:

The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

Step-by-step explanation:

An automobile manufacturer has given its van a 59.5 miles/gallon (MPG) rating. An independent testing firm has been contracted to test the actual MPG for this van since it is believed that the van has an incorrect manufacturer's MPG rating:

At the null hypothesis, we test if the mean is the same, that is:

H_0: \mu = 59.5

At the alternate hypothesis, we test that it is different, that is:

H_a: \mu \neq 59.5

The test statistic is:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

In which X is the sample mean, \mu is the value tested at the null hypothesis, \sigma is the standard deviation and n is the size of the sample.

59.5 is tested at the null hypothesis:

This means that \mu = 59.5

After testing 250 vans, they found a mean MPG of 59.2. Assume the population standard deviation is known to be 1.9.

This means that n = 250, X = 59.2, \sigma = 1.9

Value of the test statistic:

z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}

z = \frac{59.2 - 59.5}{\frac{1.9}{\sqrt{250}}}

z = -2.5

Pvalue of the test and decision:

The pvalue of the test is the probability of finding a mean that differs from 59.5 by at least 0.3, which is P(|Z|>-2.5), which is 2 multiplied by the pvalue of Z = -2.5.

Looking at the z-table, Z = -2.5 has a pvalue of 0.0062

2*0.0062 = 0.0124

The pvalue of the test is 0.0124 < 0.1, which means that there is sufficient evidence at the 0.1 level to support the testing firm's claim.

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3 years ago
1.) A florist sells bouquets for $12.25 each. Amiyah
deff fn [24]

Answer:

Step-by-step explanation:

J

6 0
3 years ago
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