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dem82 [27]
1 year ago
7

Can someone pls help me?

Mathematics
1 answer:
kirza4 [7]1 year ago
5 0

Answer:

\frac{8}{24}

Step-by-step explanation:

If Adam's family uses 8lb of apples to make pie. We know that 3 x 8 = 24, so it's probably approximately 3 out of 18.

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Solve for d.<br> 3d - d - 1 = 5<br><br> HELP
Alina [70]

Answer:

d=3

Step-by-step explanation:

4 0
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
Jarrett's puppy weighed 3 3/4 ounces at Birth. At 1 week old, the puppy weighed 5 1/8 ounces. At 2 weeks old, the puppy weighed
Ad libitum [116K]

Answer:  The answer is 7\dfrac{7}{8} ounces.

Step-by-step explanation: Given in the question the weight of Jarrett's puppy at birth, one week old and 2 weeks old are as follows:

b_0=3\dfrac{3}{4}=\dfrac{15}{4}~\textup{ounces},\\\\\\b_1=5\dfrac{1}{8}=\dfrac{41}{8}~\textup{ounces},\\\\\\b_3=6\dfrac{1}{2}=\dfrac{13}{2}~\textup{ounces}.

We can see that

b_1-b_0=b_2-b_1=\dfrac{11}{8}.

So, the weight of the baby will make an arithmetic progression with first term 'a' and common difference 'd' as follows:

b_0=\dfrac{15}{4},~~d=\dfrac{11}{8}.

Thus, the weight of the puupy after 3 weeks will be

b_3=b_0+3\times d=\dfrac{15}{4}+3\times\dfrac{11}{8}=\dfrac{30+33}{8}=\dfrac{63}{8}=7\dfrac{7}{8}~\textup{ounces}.

5 0
3 years ago
Read 2 more answers
Represent the following sentence as an algebraic expression, where “a number” is the letter x. 1 less than a number
nikitadnepr [17]

Answer:

x-1

Step-by-step explanation:

if "a number" is x, then one less than that would be subtracting one.

= x-1

Hope this helps!

3 0
3 years ago
Simplify the product using distributive property (3p+2)(5p-1)
solniwko [45]
(3p+2)(5p-1)
=3p(5p-1)+2(5p-1)
=15p^2-3p+10p-2
=15p^2+7p-2
=C
7 0
2 years ago
Read 2 more answers
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