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sweet [91]
3 years ago
5

Brand X sells 21 oz. Bags of mixed nuts that contain 29% peanuts. To make their product they combine brand A mixed nuts which co

ntain 35% peanuts and brand B mixed nuts which contain 25% peanuts. How much of each do they need to use
Mathematics
1 answer:
masha68 [24]3 years ago
4 0

Answer:

Brand X has to mix 8.4 oz of brand A mixed nuts which contain 35% peanuts and 12.6 oz of brand B mixed nuts which contain 25%, to obtain 21 oz. Bags of mixed nuts that contain 29% peanuts.

Step-by-step explanation:

Hi

We define

B_{A}=1oz @ 35\%, wich means Brand A has 1 oz containing 35% peanuts.

B_{B}=1oz @ 25\%, wich means Brand b has 1 oz containing 25% peanuts.

So we can build an equiation system

(1) B_{a}+B_{b}=21

(2)0.35 \times B_{a}+0.25 \times B_{b}=0.29 \times 21, after fixing it a little

(2)0.35B_{a}+0.25B_{b}=6.09

As we can use any method to solve the system, I used a calculator wich thrown the following results B_{a}=8.4 \ oz and B_{b}=12.6 \ oz.

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Answer:

See below for answers and explanations

Step-by-step explanation:

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<u>Problem 2:</u>

We are selecting a spade and then another spade while NOT replacing the first spade (remember that these events are independent of each other also). This means that the total card count will change by picking up the second card. Therefore, the probability of selecting a spade, followed by another spade, is (13/52)(12/51) = 156/2652 = 1/17.

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3 years ago
You deposit $1500 in a stock account. The account starts losing 2.6% interest annually. How must money do you have after 2 years
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Step-by-step explanation:

Initially in the acount there is $1500

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The equation can be writen as:

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Answer:

Step-by-step explanation:

Hello!

The given data corresponds to the variables

Y:  Annual Maintenance  Expense ($100s)

X: Weekly Usage  (hours)

n= 10

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

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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

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^Y= a + bX

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b)

H₀: β = 0

H₁: β ≠ 0

α:0.05

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

F= 47.62

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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.

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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:

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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.

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