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MatroZZZ [7]
3 years ago
14

Fast Freddie is a widget assembler. He is paid $3.27 for every widget he assembles. What is Freddie's total pay for a week in wh

ich he completes 209 widgets?
Mathematics
2 answers:
saul85 [17]3 years ago
8 0

Answer:

$683.43.

Step-by-step explanation:

We have been given that Fast Fast Freddie is a widget assembler. He is paid $3.27 for every widget he assembles.

To find Freddie's total pay we need to multiply the total number of widgets assembled in the week by the amount paid to Freddie for each widget he assembles.

\text{Freddie's total pay for the week}=\$3.27\times 209

\text{Freddie's total pay for the week}=\$683.43

Therefore, Freddie's total pay for the week is $683.43.

Brums [2.3K]3 years ago
3 0
If he is paid 3.27 for every widget assembled...and he assembles 209...then his total pay for the week is : 3.27 * 209 = $ 683.43 <==
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I’m not sure how to do this please help
Nikolay [14]

Answer:

B :step 2   she didnt collect all the like terms and calculate

Step-by-step explanation:

first rewrite  remove all (  )Parentheses

2nd   collect all like terms and calculate

a^4 + 7a -16 -12^a^3 + 5a -3

a^4 + 12a - 19 - 12a^3     ( Like terms are 7a +5a  and -16 +-3)

so she skipped the second step

3 0
3 years ago
chantal volunteered to bring 1 gallon of freshly squeezed orange juice to a picnic.the local grocery store sells 4-pound bags of
pochemuha
(128 oz) * (2 oranges)/(5 oz) * (1 bag)/(11.5 oranges) * $3.49/bag ≈ $15.54

Note that we have assumed 11.5 oranges per bag, and that partial bags are available.

If Chantal must buy whole bags of oranges, 5 bags will be needed and the cost will be $17.45.
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3 years ago
Jensen Tire &amp; Auto is in the process of deciding whether to purchase a maintenance contract for its new computer wheel align
sasho [114]

Answer:

Step-by-step explanation:

Hello!

The given data corresponds to the variables

Y:  Annual Maintenance  Expense ($100s)

X: Weekly Usage  (hours)

n= 10

∑X= 253; ∑X²= 7347; \frac{}{X}= ∑X/n= 253/10= 25.3 Hours

∑Y= 346.50; ∑Y²= 13010.75; \frac{}{Y}= ∑Y/n= 346.50/10= 34.65 $100s

∑XY= 9668.5

a)

To estimate the slope and y-intercept you have to apply the following formulas:

b= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{sumX^2-\frac{(sumX)^2}{n} } = \frac{9668.5-\frac{253*346.5}{10} }{7347-\frac{(253)^2}{10} }= 0.95

a= \frac{}{Y} -b\frac{}{X} = 34.65-0.95*25.3= 10.53

^Y= a + bX

^Y= 10.53 + 0.95X

b)

H₀: β = 0

H₁: β ≠ 0

α:0.05

F= \frac{MS_{Reg}}{MS_{Error}} ~~F_{Df_{Reg}; Df_{Error}}

F= 47.62

p-value: 0.0001

To decide using the p-value you have to compare it against the level of significance:

If p-value ≤ α, reject the null hypothesis.

If p-value > α, do not reject the null hypothesis.

The decision is to reject the null hypothesis.

At a 5% significance level you can conclude that the average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

b= 0.95 $100s/hours is the variation of the estimated average annual maintenance expense of the computer wheel alignment and balancing machine is modified when the weekly usage increases one hour.

a= 10.53 $ 100s is the value of the average annual maintenance expense of the computer wheel alignment and balancing machine when the weekly usage is zero.

c)

The value that determines the % of the variability of the dependent variable that is explained by the response variable is the coefficient of determination. You can calculate it manually using the formula:

R^2 = \frac{b^2[sumX^2-\frac{(sumX)^2}{n} ]}{[sumY^2-\frac{(sumY)^2}{n} ]} = \frac{0.95^2[7347-\frac{(253)^2}{10} ]}{[13010.75-\frac{(346.50)^2}{10} ]} = 0.86

This means that 86% of the variability of the annual maintenance expense of the computer wheel alignment and balancing machine is explained by the weekly usage under the estimated model ^Y= 10.53 + 0.95X

d)

Without usage, you'd expect the annual maintenance expense to be $1053

If used 100 hours weekly the expected maintenance expense will be 10.53+0.95*100= 105.53 $100s⇒ $10553

If used 1000 hours weekly the expected maintenance expense will be $96053

It is recommendable to purchase the contract only if the weekly usage of the computer is greater than 100 hours weekly.

4 0
3 years ago
A factory packed 9600kg of flour in to small bags and large bags. Each small bag contained 3kg of flour and each large bag conta
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Answer:

696 large bags of flour were packed.

Step-by-step explanation:

Let large bags of flour be 'y'

Let small bags of flour be 'x'. That is x=2040 (according to the question)

(3 \times x) + (5 \times y) = 9600

(3 \times 2040) + ( 5 \times y) = 9600

(6120) + (5 \times x) = 9600

(5 \times x) = 9600 - 6120

(5 \times x) = 3480

x = 3480  \div 5

x = 696

HOPE IT HELPS

#PEACE

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3 years ago
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the list S describes all possible outcomes of the event.

the fourth option

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3 years ago
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