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1 year ago

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An entry level web developer's annual pay is $55,640 based on 52 weeks per year. Due to the economy, his company is having to cu

t back on the number of weeks that it employs its web developers. If the firm cuts the work year to 49 weeks but keeps the same rate of pay, how much should the web developer expect his annual pay to decrease? Round your answer to the nearest dollar.

1 answer:

katen-ka-za [31]1 year ago

5 0

Answer:

$Reduction = \$3300$

Step-by-step explanation:

Given

$Pay\ for\ 52\ weeks = \$55,640$

Required

Determine the reduction when paid is reduced to 49 weeks

First, we need to determine the weekly pay

$Weekly\ Pay = \frac{\$55,640}{52}$

$Weekly\ Pay = \$1070$

Next, is to determine the pay for 49 weeks;

$Pay = Weekly\ Pay * 49$

$Pay = \$1070 * 49$

$Pay = \$52430$

Subtract the 49 week pay from 52 weeks pay to get the payment reduction;

$Reduction = \$55640 - \$52340$

$Reduction = \$3300$

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Write three numerical expressions that are equivalent to (0.0004) . (0.005)

yKpoI14uk [10]

<em><u>The three equivalent numerical expressions are:</u></em>

$(0.0004) . (0.005) = (\frac{4}{10000}).(\frac{5}{1000})$

$(0.0004) . (0.005)= (\frac{1}{2500}).(\frac{1}{200})$

$(0.0004) . (0.005) = (4 \times 10^{-4}).(5 \times 10^{-3})$

<em><u>Solution:</u></em>

We have to write three numerical expressions that are equivalent to (0.0004) . (0.005)

<em><u>Given expression is:</u></em>

(0.0004) . (0.005)

<em><u>First numerical expression:</u></em>

We know that 0.0004 can be written in fraction as:

$0.0004 = \frac{0.0004 \times 10000}{10000} = \frac{4}{10000}$

Similarly,

$0.005 = \frac{0.005 \times 1000}{1000} = \frac{5}{1000}$

Therefore,

$(0.0004) . (0.005) = (\frac{4}{10000}).(\frac{5}{1000})$

<em><u>Second numerical expression:</u></em>

The first numerical expression can be simplified to get second numerical expression

$(\frac{4}{10000}).(\frac{5}{1000}) = (\frac{1}{2500}).(\frac{1}{200})$

<em><u>Third numerical expression:</u></em>

The given expression can be expressed in scientific notation:

<em><u>Steps for converting to scientific notation:</u></em>

Move the decimal point in your number until there is only one non-zero digit to the left of the decimal point. The resulting decimal number is a.

Count how many places you moved the decimal point. This number is b.

If you moved the decimal to the left b is positive.

If you moved the decimal to the right b is negative.

If you did not need to move the decimal b = 0.

Write your scientific notation number as a x 10^b and read it as "a times 10 to the power of b."

Remove trailing 0's only if they were originally to the left of the decimal point.

$0.0004 = 4 \times 10^{-4}\\\\0.005 = 5 \times 10^{-3}$

Therefore, third equivalent numerical expression is:

$(0.0004) . (0.005) = (4 \times 10^{-4}).(5 \times 10^{-3})$

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1 year ago

A lightbulb manufacturer ships large consignments of lightbulbs to big industrial users. When the production process is function

Mashutka [201]

Answer:

a) P[ Pf | Ptd ] = 0,428

b) P[ Pf | Nd ] = 0,692

Step-by-step explanation:

We will call:

Pf Probability of functioning correctly 60 % or 0.6

Pfn Probability of malfunctioning 40 % or 0,4

PNd Probability of nondefective bulb

D₁ Probability of defective bulb when the process is correctly working

D₂ Probability of defective bulb when the process is in malfunction condition

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Then applying theorem´s Bayes have

P [ A | B ] = P(A) * P [ B | A ] / P(B)

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The total probability of defective bulbs is equal to, probabilty of defective bulbs when the process is Ok, plus probability of defective bulbs when the process is in malfunctioning condition, therefore

Ptd = (0,6)*(0,25) + (0,40)*(0,50) = 0,15 * 0,20 = 0,35

P[ Pf | Ptd ] = Pf * P [ Pd | Pf ] / Ptd = 0,6 * 0,25 / 0,35

P[ Pf | Ptd ] = 0,428

b) P [ Pf | Nd ]

P[ Pf | Nd ] = Pf * P [Nd | Pf ] / PNd

PNd = 1 - Ptd = 1 - 0,35 = 0,65

P[ Pf | Nd ] = 0,6 * 0,75 / 0,65 = 0,692

P[ Pf | Nd ] = 0,692

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1 year ago

Which statement best describes how to determine whether f(x) = 9 – 4x2 is an odd function? Determine whether 9 – 4(–x)2 is equiv

katovenus [111]

We have a property for odd functions, which is given below. Let f(x) be an odd function then it must satisfy the below - mentioned property.

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For this function to be odd, it must satisfy the above written property.

Replace x with -x, we get

$f(-x)=9-4(-x)^2$

And, we have to also find

$-f(x)=-(9-4x^2)$

Hence, in order to the given function to be an odd function, we must determine whether 9-4(-x)^2 is equivalent to -(9-4x^2) or not.

Therefore, C is the correct option.

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alexdok [17]

Answer:

Step-by-step explanation:

For us to be able to determine the polynomials that are divisible by (x-1), this means that x-1 must be a factor for the functon to be able to divide any of the polynimial.

Since x-1 is a factor, we can get the value of x

x-1 = 0

x =0+1

x = 1

Next is for to substitute x - 1 into the polynomial and see the ones that will give us zero

For A(x)=3x^3+2x^2-x

A(1) = 3(1)^3+2(1)^2-(1)

A(1) = 3+2-(1)

A(1) = 5-1

A(1) = 4

Since A(1) ≠ 0, then x-1 is not divisible by the polynomial function.

<u>For B(x)=5x^3-4x^2-x</u>

B(1)=5(1)^3-4(1)^2-1

B(1)=5-4-1

B(1)=1-1 = 0

Since B(1) = 0, hence x-1 is divisible by 5x^3-4x^2-x

For the polynomial C(x)= 2x^3-3x^2+2x-1

C(1)=2(1)^3-3(1)^2+2(1)-1

C(1)=2-3+2-1

C(1)= -1+1

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<u />

<u>F</u>or the polynomial D(x)=x^3+2x^2+3x+2

D(1)=1^3+2(1)^2+3(1)+2

D(1)=1+2+3+2

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2 years ago

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GuDViN [60]

Answer:

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Step-by-step explanation:

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.

In this problem, we have that:

$\mu = 109.72, \sigma = 24$

What proportion of bills are greater than $131

This proportion is 1 subtracted by the pvalue of Z when X = 131. So

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

$Z = \frac{131 - 109.72}{24}$

$Z = 0.89$

$Z = 0.89$ has a pvalue of 0.8133

1 - 0.8133 = 0.1867

18.67% of bills are greater than $131

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