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xxTIMURxx [149]
3 years ago
11

The ratio of the lengths of corresponding parts in two similar solids is 4:1.

Mathematics
1 answer:
Rina8888 [55]3 years ago
7 0

Answer:

the ratio is 16:1

Step-by-step explanation:

surface area is measured in square units, so you should square both numbers in the linear ratio to get the ratio for the surface area.

4^{2} : 1^{2} -->16:1

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What is the least common denominator for adding the fractions 4/15, 1/12, and 3/8
il63 [147K]
I think you want the least common multipule (exg LCM of 2,3,4 is 12)

so we find the factors and include all of them so like this

15=3 times 5
12=2 times 2 times 3
8=2 times 2 times 2

so to include all of them we include 3 2's, 1 3 and 1 5 or 2 times 2 times 2 times 3 times 5 or 8 times 15 or 4 times 30 or 120

the LCM is 120

so 4/15=32/120
1/12=10/120
3/8=45/120

add 32/120+10/120+45/120=(32+10+45)/120=87/120
5 0
3 years ago
Read 2 more answers
Given that 6 x − 7 y = 22 Find y when x = − 1
kykrilka [37]

Answer

y= 4

Step-by-step explanation:

6x-7y=22

6x(-1)-7y=22

-6-7y=22

-7y=22+6

-7y=28

y=-4

7 0
2 years ago
How do i answer these? can somebody try to explain it ? maybe using a photo as well
Travka [436]

\large{\color{red}{\rightarrow{-\:0.2}}}

3 0
2 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
Simplify 3(x + 4) + 5x. Write your answer in factored form.​
Anvisha [2.4K]

Answer:

8x + 12

Step-by-step explanation:

3(x + 4) + 5x

3x + 12 + 5x

8x + 12

7 0
2 years ago
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