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GREYUIT [131]
3 years ago
9

The vendor of a coffee cart mixes coffee beans that cost $9 per pound with coffee beans that cost $5 per pound. How many pounds

of each should be used to make a 75-pound blend that sells for $6.00 per pound
Mathematics
1 answer:
lidiya [134]3 years ago
8 0

Answer:

56.25 pound of the coffee that costs $5 per pound is needed

18.75 pound of the coffee that costs $9 per pound is needed

Step-by-step explanation:

Let the number of pounds be x and y respectively

The total pounds is 75;

So;

x + y. = 75 •••••••(i)

Total cost of first type

9 * x = $9x

Total cost of second type;

5 * y= $5y

75 pound at $6 per pound; total cost of this is;

6 * 75 = $450

Thus;

9x + 5y = 450 ••••••••(ii)

From i, x = 75-y

Put this into ii

9(75-y) + 5y = 450

675 -9y + 5y = 450

4y = 675-450

4y = 225

y = 225/4

y = 56.25

x = 75 - y from i

x = 75-56.25

x = $18.75

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One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0
3 years ago
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