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NISA [10]
3 years ago
10

Roger goes to a restaurant with Cathy. They order an appetizer of calamari for $8. Roger has a glass of red wine for $4 and Cath

y some white wine for $5. They both order the steak entree for $26 each. The waitress did an excellent job, so Roger and Cathy wanted to tip her 20%. Taxes are included in each menu option. Afterwards, they went to the local ice cream shop and paid $7.56. What tip did they leave the waitress? Do not include $ in answer and round to the nearest tenth. Example, if answer is $19.897, put 19.9.
Mathematics
1 answer:
Schach [20]3 years ago
8 0

Answer:

13.8

Step-by-step explanation:

8+4+5=17

26×2=52

17+52=69

69×.20=13.8

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Mr. Smith is planting a class garden for his third grade class. He surveys the class and determines that 65% prefer planting cor
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No, because he only surveyed his class which 65% of them prefer planting corn over tomatoes. In which it does not mean that the school's third grade population that prefers planting corn over tomatoes is 65%
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5.5% tax on a $309.95 purchase
oksano4ka [1.4K]

To find a 5.5% tax on $309.95, you need to multiply them together.

5.5% x 309.9

= 0.55 x 309.9

= 170.445

≈ $170.45

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F(x) = x5 + 5x4 - 5x3 - 25x2 + 4x + 20
Pie

Answer:

-4(5-x)

Step-by-step explanation:

6 0
2 years ago
At Judith's Swimwear, 75% of the 72 swimsuit styles are bikinis. How many bikini styles are
poizon [28]

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54

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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
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