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sergey [27]
2 years ago
8

Find the value of the unknown side length for the given perimeter P. P = 14 m A. 3 m B. 4 m C. 14 m D. 25 m

Mathematics
1 answer:
koban [17]2 years ago
5 0
If the perimeter is 14. Add 2,4, and 5 which is 11. Then find out how many meters does it take to get from 11 to 14. In this case the answer is 3
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Help anyone? Brainliest and five stars if correct
Leokris [45]

Answer:

X=1

Step-by-step explanation:

6x+2=9x-1

-6x on both sides

2=3x-1

+1 on both sides

3=3x

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1=x

Flip

X=1

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2 years ago
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Angles in a triangle: I did common like terms: 2x+190=180. I solved it and got -10. I checked my answer by changing x to -10. I
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7 0
3 years ago
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Mikaela friends brought a big box of crackers to the movies. 53 crackers are whole wheat and 29 are cheese flavored. The remaini
Elena-2011 [213]

Answer: 123 crackers

Step-by-step explanation:

If 1/3 of the box contains sesame seed crackers, then, 2/3 of the box contains whole wheat and cheese flavored crackers.

Total number of whole wheat and cheese flavored crackers = 53 + 29 = 82

Let x be the total number of crackers in the box, so we will have the equation

2/3 x= 82

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X =82 x 3/2

X = 123 crackers

8 0
3 years ago
Solve: 2x-10= 40<br>Solve for X​
Arturiano [62]
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2 years ago
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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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