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matrenka [14]
3 years ago
6

A group of friends wants to go to the amusement park. They have no more than $125 to spend on parking and admission. Parking is

$15, and tickets cost $12.50 per person, including tax. Write and solve an inequality which can be used to determine pp, the number of people who can go to the amusement park.
Mathematics
1 answer:
lisov135 [29]3 years ago
8 0

Answer:

100

Step-by-step explanation:

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An airplane takes off from the ground and reaches a height of 500 feet after flying 2 miles. given the formula h = d tan θ, wher
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Height covered by the plane=h= 500 feet

Distance covered by the plane along the ground=d= 2 mile

Now, convert the mile into feet,

1 mile = 5280 feet

2 mile = 10560 feet

Now, use the formula

h= d tanθ

tanθ= h/d

tanθ= 500/10560

tanθ= 25/528

Take the tan⁻¹ on both side,

θ= tan⁻¹(25/528)

θ =2.71°



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A simple random sample of 60 households in city 1 is taken. In the sample, there are 45 households that decorate their houses wi
murzikaleks [220]

Answer:

The calculated value of z = - 0.197  falls in the critical region therefore we reject the null hypothesis and conclude that  at  the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0: p1= p2 there is no difference in population proportions of households that decorate their houses with lights for the holidays

against Ha : p1≠ p2  (claim)  ( two sided)

The significance level is set at ∝= 0.05

The critical value for two tailed test at alpha=0.05 is ± 1.96

or Z∝= 0.05/2= ± 1.96

The test statistic is

Z = p1-p2/√pq(1/n1 +1/n2)

p1= proportions of households  decorating in city 1  = 45/60=0.75

p2= proportions of households  decorating in city 2 = 40/50= 0.8

p = the common proportion on the assumption that the two proportion are same.

p =   \frac{n_1p_1 +n_2p_2}{n_1+n_2}

Calculating

p =60 (0.75) + 50 (0.8) / 110

p=  45+ 40/110= 85/110 = 0.772

so  q = 1-p= 1- 0.772= 0.227

Putting the values in the test statistic and calculating

z= 0.75- 0.8/ √0.772*0.227( 1/60 + 1/50)

z= -0.05/√ 0.175244 ( 110/300)

z= -0.05/0.25348

z= -0.197

The calculated value of z = - 0.197  falls in the critical region therefore we reject the null hypothesis and conclude that  at  the 5% significance level there is significant difference in population proportions of households that decorate their houses with lights for the holidays

4 0
3 years ago
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