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grandymaker [24]

2 years ago

7

Irving Manz works for Ben’s Food Mart. He is paid weekly on the basis of a 40-hour week at the rate of $8.95 an hour with time-a

nd-a-half for overtime. During one week he works 49.5 hours. What was Irving’s time-and-a-half pay rate an hour?

2 answers:

mart [117]2 years ago

6 0

Answer: Irving earned about $363.00 during the week.

Step-by-step explanation:

AlekseyPX2 years ago

5 0

Answer:

Irving earned about $363.00 during the week.

Step-by-step explanation:

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The triangular bases of a triangular prism have three congruent sides, each measuring 10 centimeters. The height of each of the

Elina [12.6K]

We know that

[surface area ]=2*[area of the base]+[perimeter of the base]*height
area of the base=10*8.7/2-----> 43.5 cm²
perimeter of the base=10*3----> 30 cm
height of the prism=15 cm
[surface area ]=2*[43.5]+[30]*15------->537 cm²

the answer is
<span>the approximate surface area of the prism is 537 cm</span>²

5 0

2 years ago

Read 2 more answers

Which algebraic expression is a polynomial? 4x2 – 3x + StartFraction 2 Over x EndFraction –6x3 + x2 – StartRoot 5 EndRoot 8x2 +

stealth61 [152]

Answer:

Step-by-step explanation:

"4x2 – 3x + StartFraction 2 Over x EndFraction" is definitely not a polynomial because one of the exponents of x is negative.

"–6x3 + x2 – StartRoot 5 EndRoot" is a polynomial because all of the exponents of x are integers 0 or greater.

The presence of any root (fractional exponents included) of the variable x automatically disqualifies such expression: it's not a polynomial.

5 0

2 years ago

Select the correct answer. Sarah wants to print copies of her artwork. At the local print shop, it costs her $1 to make 5 copies

Ber [7]

$1 = 5copies means

$5 = 25 copies obviously

then

$x = 100 copies

100 / 5 = $x

so she needs

$20

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2 years ago

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Ashley plants a tree 45cm tall it grows by 10 cm each year. Find the amount ashley gets in the 5th year?

Genrish500 [490]

Answer:

95cm

Step-by-step explanation:

45cm is the initial height

each year grow 10cm

growing time 5 years

45+(10 × 5)=95 cm

4 0

2 years ago

A machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%). Prior to shipm

AURORKA [14]

Answer:

(a) 0.0686

(b) 0.9984

(c) 0.0016

Step-by-step explanation:

Given that a machine produces parts that are either defect free (90%), slightly defective (3%), or obviously defective (7%).

Let A, B, and C be the events of defect-free, slightly defective, and the defective parts produced by the machine.

So, from the given data:

P(A)=0.90, P(B)=0.03, and P(C)=0.07.

Let E be the event that the part is disregarded by the inspection machine.

As a part is incorrectly identified as defective and discarded 2% of the time that a defect free part is input.

So, $P\left(\frac{E}{A}\right)=0.02$

Now, from the conditional probability,

$P\left(\frac{E}{A}\right)=\frac{P(E\cap A)}{P(A)}$

$\Rightarrow P(E\cap A)=P\left(\frac{E}{A}\right)\times P(A)$

$\Rightarrow P(E\cap A)=0.02\times 0.90=0.018\cdots(i)$

This is the probability of disregarding the defect-free parts by inspection machine.

Similarly,

$P\left(\frac{E}{A}\right)=0.40$

and $\Rightarrow P(E\cap B)=0.40\times 0.03=0.012\cdots(ii)$

This is the probability of disregarding the partially defective parts by inspection machine.

$P\left(\frac{E}{A}\right)=0.98$

and $\Rightarrow P(E\cap C)=0.98\times 0.07=0.0686\cdots(iii)$

This is the probability of disregarding the defective parts by inspection machine.

(a) The total probability that a part is marked as defective and discarded by the automatic inspection machine

$=P(E\cap C)$

$= 0.0686$ [from equation (iii)]

(b) The total probability that the parts produced get disregarded by the inspection machine,

$P(E)=P(E\cap A)+P(E\cap B)+P(E\cap C)$

$\Rightarrow P(E)=0.018+0.012+0.0686$

$\Rightarrow P(E)=0.0986$

So, the total probability that the part produced get shipped

$=1-P(E)=1-0.0986=0.9014$

The probability that the part is good (either defect free or slightly defective)

$=\left(P(A)-P(E\cap A)\right)+\left(P(B)-P(E\cap B)\right)$

$=(0.9-0.018)+(0.03-0.012)$

$=0.9$

So, the probability that a part is 'good' (either defect free or slightly defective) given that it makes it through the inspection machine and gets shipped

$=\frac{\text{Probabilily that shipped part is 'good'}}{\text{Probability of total shipped parts}}$

$=\frac{0.9}{0.9014}$

$=0.9984$

(c) The probability that the 'bad' (defective} parts get shipped

=1- the probability that the 'good' parts get shipped

=1-0.9984

=0.0016

5 0

2 years ago