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tresset_1 [31]
3 years ago
13

Francis teaches the piano. He charges each student an enrollment fee of $100 plus $15 per hour of piano lessons. The average cos

t of lessons for a student per hour is $25.
If h represents the number of hours a student spends in lessons, which equation can be used to find the value of h?
Mathematics
1 answer:
Kryger [21]3 years ago
7 0

Answer:

y=15x+100

Step-by-step explanation:

increases by 15, 100 is starting free

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Answer:

2x^2+7x-4

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The cooking compartment of a microwave oven measures 15 inches long, 14 inches wide, and 9 inches high. What is the volume of th
NemiM [27]
So,

All we have to do is multiply the dimensions given.
15 * 14 * 9 = 210 * 9 = 1890

The volume of the cooking compartment of the oven is 1890 in.³
4 0
3 years ago
HOW MANY SOLUTIONS DOES<br> THIS SYSTEN OF EQUATIONS HAVE<br><br> y = 2x + 2<br> y = 2x + 2
kherson [118]

Answer: Infinitely many solutions.

Step-by-step explanation:

7 0
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• A researcher claims that less than 40% of U.S. cell phone owners use their phone for most of their online browsing. In a rando
antiseptic1488 [7]

Answer:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p > α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

Step-by-step explanation:

Set up hypotheses:

Null hypotheses = H₀: p = 0.40

Alternate hypotheses = H₁: p < 0.40

Determine the level of significance and Z-score:

Given level of significance = 1% = 0.01

Since it is a lower tailed test,

Z-score = -2.33 (lower tailed)

Determine type of test:

Since the alternate hypothesis states that less than 40% of U.S. cell phone owners use their phone for most of their online browsing, therefore we will use a lower tailed test.

Select the test statistic:  

Since the sample size is quite large (n > 30) therefore, we will use Z-distribution.

Set up decision rule:

Since it is a lower tailed test, using a Z statistic at a significance level of 1%

We Reject H₀ if Z < -1.645

We Reject H₀ if p ≤ α

Compute the test statistic:

$ Z =  \frac{\hat{p} - p}{ \sqrt{\frac{p(1-p)}{n} }}  $

$ Z =  \frac{0.31 - 0.40}{ \sqrt{\frac{0.40(1-0.40)}{100} }}  $

$ Z =  \frac{- 0.09}{ 0.048989 }  $

Z = - 1.84

From the z-table, the p-value corresponding to the test statistic -1.84 is

p = 0.03288

Conclusion:

We failed to reject H₀

Z > -1.645

-1.84 > -1.645

We failed to reject H₀

p >  α

0.03 > 0.01

We do not have significant evidence at a 1% significance level to claim that less than 40% of U.S. cell phone owners use their phones for most of their online browsing.

8 0
3 years ago
9(3+c+4) simply the expression pls help giving Brainliest
Alex
That would be 9c+63 i’m pretty sure
8 0
3 years ago
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