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netineya [11]
3 years ago
6

hurry xdd A store had apples on sale for $1.20 a pound. Joan spent $5.28 on apples. How many pounds did she buy?

Mathematics
2 answers:
hram777 [196]3 years ago
6 0

Answer:

Its 4.4 pounds.

Step-by-step explanation: To find the answer you would just divide the amount she spent and the price per pound.

Evgen [1.6K]3 years ago
5 0

Answer:

4-6 not sure where but around 4-6 maybe 5 pounds

Step-by-step explanation:

hope this helps

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If Z= 2+2i, what is z^4?
Dimas [21]

Answer:

assume z is equal to 5.

log(z^4/5) equals log(5^4/5) equals log(5^3) equals 2.096910013.

4 * log(5) - log(5) equals 2.096910013.

they give you the same answer.

this means they're equivalent.

Step-by-step explanation:

7 0
3 years ago
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Two interior angles of a triangle each measure 34 what is the measure of the third interior angel?
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3 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

5 0
3 years ago
Dont Need Help Just Trying To Give Points 20+20=????
Elenna [48]
Thanks Lots! You're awesome
4 0
3 years ago
A new car is purchased for 23900 dollars. The value of the car depreciates at 14.25% per year. To the nearest tenth of a year, h
svp [43]

Answer:

The value of the car will be 6100 dollars in 8.8 years

Step-by-step explanation:

Present value of car = $23900

The value of the car depreciates at 14.25% per year

Let x be no. of years in which  the value of the car becomes 6100 dollars

Formula: N(t)=N_0(1-r)^t

Substitute the values :

6100=23900(1-\frac{14.25}{100})^x\\\frac{6100}{23900}=(1-\frac{14.25}{100})^x\\x=8.8

Hence the value of the car will be 6100 dollars in 8.8 years

7 0
4 years ago
Read 2 more answers
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