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Genrish500 [490]
4 years ago
9

What is 2.8 divides by 0.4? Please Show your work.

Mathematics
2 answers:
Lady_Fox [76]4 years ago
8 0
It is 7 no work to show it’s easy
Eva8 [605]4 years ago
6 0
The answer is 7, work shown here

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In a carnival drawing, a green ticket wins $1, a yellow ticket wins $5, and a blue ticket wins $10. There are 100 green tickets,
Evgesh-ka [11]

The odds in favor of winning $5 or more is 3 : 13

<h3>What are the odds in favor of winning $5 or more?</h3>

The odds of winning $5 or more can be determined by finding the ratio of the total value of tickets that have a value of $5 or greater to the total value of tickets.

Total number  of the tickets that have a value of $5 or more = 25 + 5 = 30

Total number of ticket  = 25 + 5 + 100 = 130

Ratio : 30 : 130

3 ; 13

To learn more about ratios, please check: brainly.com/question/9194979

#SPJ1

5 0
2 years ago
You deposit $10,000 in an account that pays 4.5% interest compounded quarterly. Find the future value after one year. ​
Dominik [7]

Answer:

11800

Step-by-step explanation:

4.5%=0.045

0.045x10,000=450

450x4(quarterly)=1800

10,000+1800= 11800

5 0
3 years ago
Read 2 more answers
Given the following image, find the angle of rotation and the number of times each figure can rotate.​
Dmitriy789 [7]

Rotated: The figure can be rotated  2  times.  Angle of Rotation:  180 ∘

7 0
3 years ago
Q and R are independent events. P(Q) = 0.4; P(Q AND R) = 0.08. Find P(R).
algol [13]
Because Q and R are independent, you have

\mathbb P(Q\cap R)=\mathbb P(Q)\times\mathbb P(R)
0.08=0.4\times\mathbb P(R)
\implies \mathbb P(R)=0.2
8 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
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