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mel-nik [20]
2 years ago
10

Melissa's Coffee Shop makes a blend that is a mixture of two types of coffee. Type A coffee costs Melissa $ 5.95 per pound, and

type B coffee costs $ 4.15 per pound. This month's blend used three times as many pounds of type B coffee as type A, for a total cost of $ 644.00 . How many pounds of type A coffee were used?
Mathematics
2 answers:
Mekhanik [1.2K]2 years ago
6 0
5.95x+4.15(3x)=644
Solve for x
X=35
Check
(5.95×35)+4.15×(3×35)=644
olya-2409 [2.1K]2 years ago
5 0
<span>B=89 lbs ANSWER: The mix contained 89 pounds of Type B coffee.
CHECK:
167lbs-89lbs=78 lbs She used 78 pounds of Type A coffee.
$4.35A+$5.70B=$846.60
$4.35(78)+$5.70(89)=$846.60
$339.30+$507.30=$846.60
$846.60=$846.60</span>
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A rectangle has an area of 22.64 inches squared. If the coordinates of one of its sides are (0, 0) and (4, 4), what are the dime
wolverine [178]

Answer:

width= 4

hieght- 5.65

Step-by-step explanation:

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which is 4 root 2

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2 years ago
Consider the accompanying data on breaking load (kg/25mm width) for various fabrics in both an unabraded condition and an abrade
WITCHER [35]

Answer:

The critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998.

The test statistic (t=1.729) is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

Step-by-step explanation:

We have a matched-pair t-test for the difference.

We have the null and alternative hypothesis written as:

H_0: \mu_d=0\\\\H_a: \mu_d>0

We have n=8 pairs of data. We calculate the difference as:

d_1=U_1-A_1=36.4-28.5=7.9

Then, with this procedure we get the sample for d:

d=[7.9,\, 35,\, 5.5,\, 4.2,\, 6.7,\, -3.7,\, -0.9,\, 3.3]

The sample mean and standard deviation are:

M=\dfrac{1}{n}\sum_{i=1}^n\,x_i\\\\\\M=\dfrac{1}{8}(7.9+35+5.5+. . .+3.3)\\\\\\M=\dfrac{58}{8}\\\\\\M=7.25\\\\\\s=\sqrt{\dfrac{1}{n-1}\sum_{i=1}^n\,(x_i-M)^2}\\\\\\s=\sqrt{\dfrac{1}{7}((7.9-7.25)^2+(35-7.25)^2+(5.5-7.25)^2+. . . +(3.3-7.25)^2)}\\\\\\s=\sqrt{\dfrac{985.08}{7}}\\\\\\s=\sqrt{140.73}=11.86\\\\\\

Now, we can perform the one-tailed hypothesis test.

The significance level is 0.01.

The sample has a size n=8.

The sample mean is M=7.25.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=11.86.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{11.86}{\sqrt{8}}=4.1931

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{7.25-0}{4.1931}=\dfrac{7.25}{4.1931}=1.729

The degrees of freedom for this sample size are:

df=n-1=8-1=7

This test is a right-tailed test, with 7 degrees of freedom and t=1.729, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t>1.729)=0.0637

As the P-value (0.0637) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.01, there is not enough evidence to support the claim that the true difference is significantly bigger than 0.

If we use the critical value approach, the critical value for a one-tailed test with 7 degrees of freedom and level of significance α=0.01 is tc=2.998. The test statistic is below the critical value and falls within the acceptance region, so it is failed to reject the null hypothesis.

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3 years ago
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malfutka [58]

Answer:

6.75 is the answer

Step-by-step explanation:

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