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nataly862011 [7]

2 years ago

11

Phil has budgeted 120 to rent a car during his weeklong vacation. At the rate of $40 per week plus 10 cents per mile how many mi

les can he drive the car and stay within the budget

1 answer:

Elanso [62]2 years ago

4 0

800 miles 120-40=80 so he has $80 worth of miles to use. He can go 10 miles on $1 so 80*10=800

He can go 800 miles.

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Samira buys a car the car loses 23% of its value in the first year. at the end of the first year her car is worth 6575.80 dollar

Andreyy89

Answer: Samira paid $8540 for her car.

Step-by-step explanation:

Let x = Amount paid by Samira for her car.

Given: Depreciation rate = 23%

Worth of car after one year = $ 6575.80

As per given,

x - 23 % of x = 6575.80

⇒ x - 0.23x = 6575.80

⇒ 0.77x= 6575.80

⇒ x= $8540 [Divide both sides by 0.77]

Hence, Samira paid $8540 for her car.

6 0

3 years ago

Suppose that the mean time that visitors stay at a museum is 94.2 minutes with a standard deviation of 15.5 minutes. The standar

Maurinko [17]

Answer:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

Step-by-step explanation:

For this case we know that mean time that visitors stay at a museum is given by:

$\bar X = 94.2$

The standard deviation is given by:

$s= 15.5$

And the standard error is given by:

$SE = \frac{s}{\sqrt{n}} =3.1$

And we want to interval captures 68% of the means for random samples of 25 scores and for this case the critical value can be founded like this using the normal standard distribution or excel:

$z_{\alpha/2}= \pm 0.994$

We can find the interval like this:

$\bar X \pm ME$

And replacing we got:

$94.2 -0.994*3.1 = 91.1186$

$94.2 +0.994*3.1 = 97.2814$

And the 68% confidence interval is given by (91.1186, 97.2814)

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2 years ago

When can you write ratios as a fraction?

k0ka [10]

Answer:

In a mathequation when needed

Step-by-step explanation:

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2 years ago

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liq [111]

Answer:

320m

Step-by-step explanation

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