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Wewaii [24]
3 years ago
11

Henry would receive $8 per hour, His total earnings amount was $1,920. How many hours did henry work for?

Mathematics
1 answer:
enot [183]3 years ago
6 0

Answer: 240

Step-by-step explanation: 1920/8

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Could you help me and answer a couple questions for me?
kari74 [83]

Answer:

I think no. <em>D</em> is the answer

5 0
3 years ago
How long will the sum of Rs 5,000 take to reach the amount RS 8750 at the rate of 15% per annum​
Nonamiya [84]

Answer:

It will take 5 yrs the sum of 5000 will take to reach Rs 8750 at the rate 15%

Step-by-step explanation:

T= ?

Amount = Rs 8750

R = 15%

P = Rs 5000

so

I = A-P

= 8750 - 5000

= Rs 3750

so

I = PTR/100

3750 = (5000*15*T)/100

or, 375000 = 75000*T

or, T = 375000/75000

so T = 5 yrs

8 0
3 years ago
(1 point) The cost of unleaded gasoline once followed an unknown distribution with a mean of $4.59 and a standard deviation of $
marin [14]

Answer:

a) unknown

b) 1.43% probability that the average price for 30 gas stations is less than $4.55.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

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

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

\mu = 4.59, \sigma = 0.1

a) What's the approximate probability that the average price for 16 gas stations is over $4.69?

Sample size less than 30 and unknow distribution. Therefore, the central limit theorem cannot be applied and the answer is unknown.

b) Find the probability that the average price for 30 gas stations is less than $4.55.

Now n = 30, s = \frac{0.1}{\sqrt{30}} = 0.0183

This probability is the pvalue of Z when X = 4.55. So

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

By the Central Limit Theorem

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

Z = \frac{4.55 - 4.59}{0.0183}

Z = -2.19

Z = -2.19 has a pvalue of 0.0143

1.43% probability that the average price for 30 gas stations is less than $4.55.

5 0
4 years ago
Help with this I don't know how to solve
miv72 [106K]

Answer:

approx 10.25

Step-by-step explanation:

According to Euclid theorem :

EG^2= DG*FG= 16*5=80

From right triangle EDG according to Phitagor's  theorem:

DE^2=EG^2+DG^2= 80+25=105

DE=sqrt(105)= approx 10.25

8 0
3 years ago
Please help asap
djverab [1.8K]

Answer:

come fast in English classhshshshdhhhdhhshdhdbb

shhshshsh

5 0
3 years ago
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