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Scrat [10]
2 years ago
6

1. John earns $10 per hour for any work up to 40 hours. For any time over 40 in a week, he receives 1.5 times his normal pay. Ho

w much would John earn for a week in which he worked 48 hours?
2. Landon earns $11.25 per hour for any work up to 40 hours. For any time over 40 in a week, he receives 1.5 times his normal pay. How much would John earn for a week in which he worked 47 hours?
Mathematics
1 answer:
yanalaym [24]2 years ago
8 0

Answer:

1. 520

2. 568.125

Step-by-step explanation:

1.

10 per 40 hours + 1.5 overtime

1.5 overtime = 10(1.5) = 15

10 (40) + 15 (8)

= 400 + 120

= 520

2.

11.25 per 40 hours + 1.5 overtime

1.5 overtime = 11.25(1.5) = 16.875

11.25 (40) + 16.875 (7)

= 450 + 118.125

= 568.125

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pentagon [3]

Answer:

Correct choce is (a). 30.

Step-by-step explanation:

Given equation is f(x)=x^2-x.

Now we need to find the value of (fof)(3). Then select correct matching choice from the given choices. Where given choices are:

(a). 30

(a). 33

(a). 6

(a). -6

Now let's find  (fof)(3)

(fof)(3)=f(f(3))

(fof)(3)=f(3^2-3)

(fof)(3)=f(9-3)

(fof)(3)=f(6)

(fof)(3)=6^2-6

(fof)(3)=36-6

(fof)(3)=30

Hence correct choce is (a). 30.

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2 years ago
Um........ Help? With step by step explanation ​
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2 years ago
Although cities encourage carpooling to reduce traffic congestion, most vehicles carry only one person. For example, 64% of vehi
ira [324]

Answer:

a) 0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

b) 0.996 is the probability that more than half of the vehicles  carry just one person.    

Step-by-step explanation:

We are given the following information:

A) Binomial distribution

We treat vehicle on road with one passenger as a success.

P(success) = 64% = 0.64

Then the number of vehicles follows a binomial distribution, where

P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}

where n is the total number of observations, x is the number of success, p is the probability of success.

Now, we are given n = 10

We have to evaluate:

P(x \geq 6) = P(x =6) +...+ P(x = 10) \\= \binom{10}{6}(0.64)^6(1-0.64)^4 +...+ \binom{10}{10}(0.64)^{10}(1-0.79)^0\\=0.7291

0.7291 is the probability that more than half out of 10 vehicles carry just 1 person.

B) By normal approximation

Sample size, n = 92

p = 0.64

\mu = np = 92(0.64) = 58.88

\sigma = \sqrt{np(1-p)} = \sqrt{92(0.64)(1-0.64)} = 4.60

We have to evaluate the probability that more than 47 cars carry just one person.

P(x \geq 47)

After continuity correction, we will evaluate

P( x \geq 46.5) = P( z > \displaystyle\frac{46.5 - 58.88}{4.60}) = P(z > -2.6913)

= 1 - P(z \leq -2.6913)

Calculation the value from standard normal z table, we have,  

P(x > 46.5) = 1 - 0.004 = 0.996 = 99.6\%

0.996 is the probability that more than half out of 92 vehicles carry just one person.

5 0
3 years ago
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