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vichka [17]
3 years ago
13

Nuts.com sells raw almonds for $10 per pound and roasted pistachios for $12 per pound. Christopher spends $70, before taxes and

shipping, to purchase 6.3 total pounds of nuts. Let a represent the number of pounds of almonds and let p represent the number of pounds of pistachios. Which system of equations represents the situation?
Question 1 options:


{10a+p=70
a+70p=6.3



{10a+12p=70
a+p=6.3



{10a+12p=6.3
a+p=70



{10a+12p=70
10a+12p=6.3
Mathematics
1 answer:
galben [10]3 years ago
3 0

Answer:

second equation

Step-by-step explanation:

10a + 12p = 70  (10 goes with almonds [a], 12 goes with pistachios[p], $10 and $12 is price per pound and $70 is total price, so that's why its = to 70)

a + p = 6.3        (this is total almonds and pistachios adding up to 6.3 pounds)

a = pounds of almonds

p = pounds of pistachios

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Answer:

a) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

b) Test statistic t=-1.565

P-value = 0.0612

NOTE: the sample size is n=65.

c) Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

d) Null and alternative hypothesis

H_0: \mu=32.48\\\\H_a:\mu< 32.48

Test statistic t=-1.565

Critical value tc=-1.669

t>tc --> Do not reject H0

Do not reject H0. We cannot conclude that the price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

Then, the null and alternative hypothesis are:

H_0: \mu=32.48\\\\H_a:\mu< 32.48

The significance level is 0.05.

The sample has a size n=65.

The sample mean is M=30.15.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=12.

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

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{12}{\sqrt{65}}=1.4884

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{30.15-32.48}{1.4884}=\dfrac{-2.33}{1.4884}=-1.565

The degrees of freedom for this sample size are:

df=n-1=65-1=64

This test is a left-tailed test, with 64 degrees of freedom and t=-1.565, so the P-value for this test is calculated as (using a t-table):

\text{P-value}=P(t

As the P-value (0.0612) is bigger than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

At a significance level of 0.05, there is not enough evidence to support the claim that the mean price in Providence for a bottle of red wine that scores 4.0 or higher on the Vivino Rating System is less than the population mean of $32.48.

<u>Critical value approach</u>

<u></u>

At a significance level of 0.05, for a left-tailed test, with 64 degrees of freedom, the critical value is t=-1.669.

As the test statistic is greater than the critical value, it falls in the acceptance region.

The null hypothesis failed to be rejected.

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