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lakkis [162]
3 years ago
9

The length of the diogonals of a rectangular garden is 34m. If it's longer side measures 30m, find the perimeter of garden. ​

Mathematics
1 answer:
MakcuM [25]3 years ago
6 0

Answer:

92 m

Step-by-step explanation:

Use the Pythagorean theorem to find the length of the shorter side.

Let w = length of the shorter side.

w^{2}  + 30^{2}  =  34^{2} \\ w^{2}  + 900 = 1156\\w^{2}  =  256\\w = \sqrt{256} = 16

The perimeter is 2l + 2w = 2(30) + 2(16) = 60 + 32 = 92 m

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13299220232+1329220232​
denpristay [2]

Answer:

14,628,440,464

Step-by-step explanation:

Perform simple addition.

Glad I was able to help!!

6 0
3 years ago
Read 2 more answers
Jason bought 12 new CDs to add to his
BabaBlast [244]

Answer:

He had origanlly had 68 altogether

Step-by-step explanation:

Because if u count the scrateched ones which was 6 is now 62 then add the other six because there was 12 and his brother scrateched half which is basically 12 / 2 = 6 so if u add the six is 68

What Im basically saying is 56 + 12 = 68

5 0
3 years ago
A box of cereal weighs 600 grams. How much is this weight in pounds. Explain or show your reasoning.
laila [671]

Answer:

1.323 pounds rounded up

Step-by-step explanation:

600 * .0022lbs + 1.32277

5 0
3 years ago
In order to estimate the difference between the average hourly wages of employees of two branches of a department store, the fol
uysha [10]

Answer:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

Step-by-step explanation:

For this problem we have the following data given:

\bar X_1 = 9 represent the sample mean for one of the departments

\bar X_2 = 8 represent the sample mean for the other department

n_1 = 25 represent the sample size for the first group

n_2 = 20 represent the sample size for the second group

s_1 = 2 represent the deviation for the first group

s_2 =1 represent the deviation for the second group

Confidence interval

The confidence interval for the difference in the true means is given by:

(\bar X_1 -\bar X_2) \pm t_{\alpha/2} \sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}

The confidence given is 95% or 9.5, then the significance level is \alpha=0.05 and \alpha/2 =0.025. The degrees of freedom are given by:

df=n_1 +n_2 -2= 20+25-2= 43

And the critical value for this case is t_{\alpha/2}=2.02

And replacing we got:

(9-8) -2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 0.0743

(9-8) +2.02 \sqrt{\frac{2^2}{25} +\frac{1^2}{20}}= 1.926

And we are 9% confidence that the true mean for the difference of the population means is given by:

0.0743 \leq \mu_1 -\mu_2 \leq 1.926

4 0
3 years ago
Please help me with the answer!! There is the picture of the problem^
Valentin [98]
(2,14) is the answer I’m pretty sure
7 0
3 years ago
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