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3 years ago

5

Which unit rate is the lowest price per ounce? Choice A: 5 ounces of raisins for $1.49 Choice B: 12 ounces of raisins for $3.59

Choice A Choice B The unit rates are equal. The unit rates cannot be determined.

2 answers:

hoa [83]3 years ago

7 0

5 oz for 1.49.....1.49/5 = 0.298 per oz
12 oz for 3.59...3.59/12 = 0.299 per oz

when rounded, the unit rates are equal

Delvig [45]3 years ago

5 0

Answer:

Choice A is the lowest price per ounce

Step-by-step explanation:

Choice A: 5 ounces of raisins for $1.49

Cost of 5 ounces of raisins = $1.49

Cost of 1 ounce of raisins = $\frac{1.49}{5}$

= $0.298$

Unit rate =$ 0.298

Choice B: 12 ounces of raisins for $3.59

Cost of 12 ounces of raisins = $3.59

Cost of 1 ounce of raisins = $\frac{3.59}{12}$

= $0.299$

Unit rate =$0.299

Hence Choice A is the lowest price per ounce

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2 years ago

Complete parts (a) through (c) below. (a) Determine the critical value(s) for a right-tailed test of a population mean at t

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a) The critical value on this case would be $t_{crit}=1.325$

b) The critical value on this case would be $t_{crit}=-1.345$

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Step-by-step explanation:

Part a

The system of hypothesis on this case would be:

Null hypothesis: $\mu \leq \mu_0$

Alternative hypothesis: $\mu > \mu_0$

Where $\mu_0$ is the value that we want to test.

In order to find the critical value we need to find first the degrees of freedom, on this case that is given df=20. Since its an upper tailed test we need to find a value a such that:

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And we can use excel in order to find this value with this function: "=T.INV(0.9,20)". The 0.9 is because we have 0.9 of the area on the left tail and 0.1 on the right.

The critical value on this case would be $t_{crit}=1.325$

Part b

The system of hypothesis on this case would be:

Null hypothesis: $\mu \geq \mu_0$

Alternative hypothesis: $\mu < \mu_0$

Where $\mu_0$ is the value that we want to test.

In order to find the critical value we need to find first the degrees of freedom, given by:

$df=n-1=15-1=14$

Since its an lower tailed test we need to find b value a such that:

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And we can use excel in order to find this value with this function: "=T.INV(0.1,14)". The 0.1 is because we have 0.1 of the area accumulated on the left of the distribution.

The critical value on this case would be $t_{crit}=-1.345$

Part c

The system of hypothesis on this case would be:

Null hypothesis: $\mu = \mu_0$

Alternative hypothesis: $\mu \neq \mu_0$

Where $\mu_0$ is the value that we want to test.

In order to find the critical value we need to find first the degrees of freedom, given by:

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Since its a two tailed test we need to find c value a such that:

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And we can use excel in order to find this value with this function: "=T.INV(0.025,11)". The 0.025 is because we have 0.025 of the area on each tail.

The critical values on this case would be $t_{crit}=\pm 2.201$

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