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omeli [17]
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.
Mathematics
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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If B is the midpoint of AC, then |AB| = |AC|.
|AB| = 3x + 2
|BC| = 5x - 10
Therefore we have the equation:
3x + 2 = 5x - 10     |subtract 2 from both sides
3x = 5x - 12      |subtract 5x from both sides
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On a trip mrs. ahmed drove 188 miles in 4 hours. On the return trip, she took a different route and traveled 197 miles in 4.5 ho
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Answer:

45.29 miles per hour

Step-by-step explanation:

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Cheers.

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Edit:  Thanks to Chegsnut36 for calculation correction

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5 0
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
olga_2 [115]

Answer:

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

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

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:

P(t_{20}>a) = 0.1

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:

P(t_{14}

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:

df=n-1=12-1=11

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

P(t_{11}>c) = 0.025 or P(t_{11}

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

5 0
3 years ago
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