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arsen [322]
2 years ago
8

By 3:00 pm, the temperature had increased by another 5%F. Was the temperature at 3:00 pm positive or negative?How do you know?Wh

at was the temperature at 3:00pm?
Mathematics
1 answer:
victus00 [196]2 years ago
4 0

Answer:

Part a) The temperature at noon was -3°F

Part b) The temperature at 3:00 pm was +2°F

Part c) The temperature at 11:00 pm was -6°F

Step-by-step explanation:

The complete question is

When Leo woke up he saw that the temperature was -8°F. By noon the temperature has increased 5°F. Part a) What was the temperature at noon?

Part b) By 3:00 pm, the temperature increased by another 5 degrees Fahrenheit was the temperature at 3:00 pm positive or negative?

Part c) By 11:00 pm, the temperature had dropped 8°F. Was the

temperature at 11:00 PM positive or negative? Explain.​

Part a) we know that

You can use a number line to adds the numbers

-8+5=-3°F

therefore

The temperature at noon was -3°F

Part b) we know that

the temperature increased by another 5 degrees

You can use a number line to adds the numbers

-3+5=+2°F

therefore

The temperature at 3:00 pm was +2°F

Part c) we know that

the temperature had dropped 8°F

You can use a number line to adds the numbers

2-8=-6°F

therefore

The temperature at 11:00 pm was -6°F

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Answer:

C & D

Step-by-step explanation:

A. = 48 cubic units

B. = 48 cubic units

C. = 64 cubic units

D. = 44 cubic units

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Answer:A unit rate is a rate with 1 in the denominator. If you have a rate, such as price per some number of items, and the quantity in the denominator is not 1, you can calculate unit rate or price per unit by completing the division operation: numerator divided by denominator.

Step-by-step explanation:

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Maureen earned some money doing odd jobs last summer and put it in a savings account that earns 9% interest compounded quarterly
ivolga24 [154]

Answer:

Maureen earned $123.19 doing odd jobs

Step-by-step explanation:

The compound interest formula is given by:

A(t) = P(1 + \frac{r}{n})^{nt}

Where A(t) is the amount of money after t years, P is the principal(the initial sum of money), r is the interest rate(as a decimal value), n is the number of times that interest is compounded per unit year and t is the time in years for which the money is invested or borrowed.

In this question:

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This is P.

A(t) = P(1 + \frac{r}{n})^{nt}

300 = P(1 + \frac{0.09}{4})^{4*10}

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3 years ago
Question Help Suppose that the lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a
koban [17]

Answer:

a)3.438% of the light bulbs will last more than 6262 hours.

b)11.31% of the light bulbs will last 5252 hours or less.

c) 23.655% of the light bulbs are going to last between 5858 and 6262 hours.

d) 0.12% of the light bulbs will last 4646 hours or less.

Step-by-step explanation:

Normally distributed problems can be solved by the z-score formula:

On a normaly distributed set with mean \mu and standard deviation \sigma, the z-score of a value X is given by:

Z = \frac{X - \mu}{\sigma}

After we find the value of Z, we look into the z-score table and find the equivalent p-value of this score. This is the probability that a score will be LOWER than the value of X.

In this problem, we have that:

The lifetimes of light bulbs are approximately normally​ distributed, with a mean of 5656 hours and a standard deviation of 333.3 hours.

So \mu = 5656, \sigma = 333.3

(a) What proportion of light bulbs will last more than 6262 ​hours?

The pvalue of the z-score of X = 6262 is the proportion of light bulbs that will last less than 6262. Subtracting 100% by this value, we find the proportion of light bulbs that will last more than 6262 hours.

Z = \frac{X - \mu}{\sigma}

Z = \frac{6262 - 5656}{333.3}

Z = 1.82

Z = 1.81 has a pvalue of .96562. This means that 96.562% of the light bulbs are going to last less than 6262 hours. So

P = 100% - 96.562% = 3.438% of the light bulbs will last more than 6262 hours.

​(b) What proportion of light bulbs will last 5252 hours or​ less?

This is the pvalue of the zscore of X = 5252

Z = \frac{X - \mu}{\sigma}

Z = \frac{5252- 5656}{333.3}

Z = -1.21

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This is the pvalue of the zscore of X = 6262 subtracted by the pvalue of the zscore X = 5858

For X = 6262, we have that Z = 1.81 with a pvalue of .96562.

For X = 5858

Z = \frac{X - \mu}{\sigma}

Z = \frac{5858- 5656}{333.3}

Z = 0.61

Z = 0.61 has a pvalue of .72907.

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P = .96562 - .72907 = .23655

23.655% of the light bulbs are going to last between 5858 and 6262 hours.

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This is the pvalue of the zscore of X = 4646

Z = \frac{X - \mu}{\sigma}

Z = \frac{4646- 5656}{333.3}

Z = -3.03

Z = -3.03 has a pvalue of .0012. This means that 0.12% of the light bulbs will last 4646 hours or less.

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I think it is C-two real solution
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