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expeople1 [14]
3 years ago
12

It is attached please help! Why is it c?

Mathematics
1 answer:
Sergeeva-Olga [200]3 years ago
6 0
The answer is B.
The total hours worked is 5 x 8 teams is 40 hours x 2 cars per hour is 80 cars.
$600/80= $7.50 per car
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Which of the following equations could be used to determine the value of X
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D

Step-by-step explanation:

If the original price is x and you deduct 3, you'l have x - 3.

If you buy 4 of them, you multiply that price by 4, giving you 4(x - 3).

Then, if that value is equal to 27, you get 4(x - 3) = 27

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the graph that goes up the most

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Super Bowl XLVI was played between the New York Giants and the New England Patriots in Indianapolis. Due to a decade-long rivalr
Diano4ka-milaya [45]

Answer:

Probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

Step-by-step explanation:

We are given that Due to a decade-long rivalry between the Patriots and the city's own team, the Colts, most Indianapolis residents were rooting heartily for the Giants. Suppose that 90% of Indianapolis residents wanted the Giants to beat the Patriots.

Let p = % of Indianapolis residents wanted the Giants to beat the Patriots = 90%

The z-score probability distribution for proportion is given by;

                   Z = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where,  \hat p = % of Indianapolis residents who were rooting for the Giants in Super Bowl XLVI in a sample of 200 residents = \frac{170}{200} = 0.85

           n = sample of residents = 200

So, probability that from a sample of 200 Indianapolis residents, fewer than 170 were rooting for the Giants in Super Bowl XLVI is given by = P(\hat p < 0.85)

     P(\hat p < 0.85) = P( \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < \frac{0.85-0.90}{\sqrt{\frac{0.85(1-0.85)}{200} } } ) = P(Z < -1.98) = 1 - P(Z \leq 1.98)

                                                                   = 1 - 0.97615 = 0.02385

<em>The above probability is calculated using z table by looking at value of x = 01.98 in the z table which have an area of 0.97615.</em>

<em />

Therefore, probability that fewer than 170 were rooting for the Giants in Super Bowl XLVI is 0.02385.

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24.9+(-51.7)= what’s the answer for this question?
Molodets [167]

Answer:−26.8

Step-by-step explanation:

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