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son4ous [18]
2 years ago
10

Boht agyi yaadehmm:)​

Mathematics
2 answers:
topjm [15]2 years ago
4 0

Answer:

is this a different language?

rodikova [14]2 years ago
3 0
I don’t know what it says but hey! :)
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5+3 = 8
8 is 100%
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Equation practice with angle addition
Triss [41]

Answer:

36 degrees

Step-by-step explanation:

since, m<AOB is a right angle, we shall just create an equation:

M<AOB+M<BOC=90 degrees

Substitute: 6z-12+3z+30=90

Solve: 6z+3x=9z

-12+30=18

so, 9z+18=90 degrees

           -18  -18

9z=72

z=8

So if z=8, then we should fit it so that it matches measure of AOB.

6z-12

6 times 8=48

48-12= 36

6 0
3 years ago
La mama de fabiana tiene un vivero. En su ultimo pedido compro 5 fresias, una con flores rojas, una con flores blancas, una con
lys-0071 [83]

The computation shows that the number of flowers will be 10.

<h3>How to calculate the value?</h3>

The information illustrates the flowers that were brought for the nursery.

Based on the information given, the total number of flowers bought for the nursery will be:

= 5 freesias + 3 rosebrushes + 2 calls lilies

= 10

In conclusion, the total number of flowers is 10.

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6 0
2 years ago
Whos ur celebrity crush?<br><br> mine is harry styles
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Answer:

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8 0
3 years ago
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Researchers interviewed street prostitutes in Canada and the United States. The mean age of the 100 Canadian prostitutes upon en
levacccp [35]

Answer:

Step-by-step explanation:

The null and the alternative hypothesis is:

H_o: \mu_c \ge \mu_{us}

H_a : \mu_c < \mu_{us}

The t- student test statistics can be computed as:

t = \dfrac{x_c- x_{us}}{\sqrt{\dfrac{\sigma_c^2}{n_c} + \dfrac{\sigma_{us}^2}{n_{us}} }}

t = \dfrac{19- 21}{\sqrt{\dfrac{7^2}{100} + \dfrac{8^2}{130} }}

t = -2.017  

degree of freedom = (n₁ - 1) + (n₂ - 1)  

= (100 - 1) + (130 - 1)

= 228

Using the data of t-value and degree of freedom;  

The P-value = -0.224

Decision rule: Do not reject the null hypothesis if the p-value is greater than ∝(0.01)

Conclusion: We reject the null hypothesis since the p-value is less than ∝.

Therefore, there is enough evidence to conclude that the mean age of entering prostitution in Canada is lower than that of the United States.

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