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MariettaO [177]
3 years ago
5

A 16-ounce carton of milk costs 63 cents, while a half-gallon jug of milk sells for $2.24. How much more expensive, in cents per

quart, is the carton of milk compared to the half-gallon jug of milk? (32 ounces = 1 quart = 1/4 gallon)
Mathematics
1 answer:
hram777 [196]3 years ago
5 0
1 quart of the 16oz jug is $1.26 (.63x2) 1 quart of the half gallon is $1.12 (2.24/2) so 1.26-1.12= 0.14 the 16oz jugs are 14 cents more expensive than the half gallon
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find the simple interest on 10000 from 1 July to 12 september 2007 at 5% per year calculate the interest​
rusak2 [61]

Answer:

Simple interest: 83.33, Final amount: 10083.33

Step-by-step explanation:

you dont get a explanation >:)

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3 years ago
Sunspots have been observed for many centuries. Records of sunspots from ancient Persian and Chinese astronomers go back thousan
Lunna [17]

Answer:

(a) Null Hypothesis, H_0 : \mu \leq 41  

    Alternate Hypothesis, H_A : \mu > 41

(b) The value of z test statistics is 1.08.

(c) We conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

Step-by-step explanation:

We are given that in a random sample of 40 such periods from Spanish colonial times, the sample mean is x¯ = 47.0. Previous studies of sunspot activity during this period indicate that σ = 35.

It is thought that for thousands of years, the mean number of sunspots per 4-week period was about µ = 41.

Let \mu = <u><em>mean sunspot activity during the Spanish colonial period.</em></u>

(a) Null Hypothesis, H_0 : \mu \leq 41     {means that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41}

Alternate Hypothesis, H_A : \mu > 41     {means that the mean sunspot activity during the Spanish colonial period was higher than 41}

The test statistics that would be used here <u>One-sample z test statistics</u> as we know about the population standard deviation;

                            T.S. =  \frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }  ~ N(0,1)

where, \bar X = sample mean = 47

            σ = population standard deviation = 35

            n = sample of periods from Spanish colonial times = 40

So, <em><u>the test statistics</u></em>  =  \frac{47-41}{\frac{35}{\sqrt{40} } }

                                      =  1.08

(b) The value of z test statistics is 1.08.

(c) <u>Now, the P-value of the test statistics is given by;</u>

                P-value = P(Z > 1.08) = 1 - P(Z < 1.08)

                              = 1 - 0.8599 = <u>0.1401</u>

Since, the P-value of the test statistics is higher than the level of significance as 0.1401 > 0.05, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that the mean sunspot activity during the Spanish colonial period was lesser than or equal to 41.

8 0
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Step-by-step explanation:

A=1/2bh

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