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eimsori
2 months ago
15

Happy is four times as old as Grumpy. The sum of their age is 100. How old are Happy and Grumpy?

Mathematics
2 answers:
PIT_PIT [12.4K]2 months ago
5 0

Answer:

Grumpy's age is 20 while Happy's age is 80.

Step-by-step explanation:

babunello [11.8K]2 months ago
3 0
Happy is 25 years old, while Grumpy is 4 years old.
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If there were 49 cars in a line that stretched 528 feet, what is the average car length? assume that the cars are lined up bumpe
AnnZ [12381]
If 49 cars are lined up and extend 528 feet, what would the average length per car be, assuming they are bumper-to-bumper? The result can be calculated as follows: we know the number of cars (49) and the total length (528 feet). Therefore, the average length is calculated as 528 feet divided by 49, which gives an average length of about 10.78 feet.
6 0
2 months ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
babunello [11817]

The 99% confidence interval for the actual mean difference between average mail-order and internet purchase amounts falls within [$(-31.82), $12.02].

Step-by-step clarification:

We know a random sample of 16 mail-order sales receipts shows a mean sale amount of $74.50 and a standard deviation of $17.25.

For internet sales, a random sample of 9 receipts gives a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal value utilized for constructing a 99% confidence interval for the true mean difference is given by;

                      P.Q.  =  

 ~

where,

= sample mean of mail-order sales = $74.50 \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

= sample mean of internet sales = $84.40

\bar X_1 = standard deviation for mail-order sales = $17.25

\bar X_2 = standard deviation for internet sales = $21.25

s_1 = number of mail-order sales receipts = 16

= number of internet sales receipts = 9s_2

Furthermore,  

 =  n_1 = 18.74

n_2

The actual mean difference between average mail-order and internet purchases is denoted by (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} }

).

Thus, the 99% confidence interval for (\mu_1-\mu_2) is expressed as;

      = \mu_1-\mu_2 Here, the t critical value at the 0.5% significance level with 23 degrees of freedom is 2.807.           =

          = [$-31.82, $12.02](\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_) \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

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

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

5 0
2 months ago
Ben has 400 counters in a bag. He gives 35 of the counters to Sonia 130 of the counters to Phil 75 of the counters to Lance What
lawyer [12517]

Answer:

2/5

Step-by-step explanation:

By subtracting 130, 35, and 74 from the total of 400, you arrive at 160 left. Next, simplify this fraction by finding the greatest common divisor to get 2/5!

8 0
3 months ago
Read 2 more answers
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