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OLEGan [10]
3 years ago
5

What is 5x-2y=4 6x+3y=-6

Mathematics
1 answer:
telo118 [61]3 years ago
5 0

Answer:

x = 0 , y = -2

Step-by-step explanation:

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Andrej [43]
Okkk .. this is what this app is for anyways to help.
5 0
3 years ago
I need help plzzzzz anyone plzz I have it due soon<br>​
KATRIN_1 [288]

Answer:

The correct answer is C

Step-by-step explanation:

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6 0
3 years ago
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Find the volume of a right circular cone that has a height of 7.4 in and a base with a diameter of 5.4 in. Round your answer to
JulijaS [17]

Answer:

approximately 56.51 cubic inches

Step-by-step explanation:

volume of a cone = 1 / 3 × pi × radius^2 × height

= 1/3 × 22/7 × 7.29 × 7.4

= 56.5148 cubic inches

8 0
2 years ago
Es urgente ¡¡¡ si la MH de a y 4 es 6 y la MH de 8 y b es 12 calcula la MH de a y b
TiliK225 [7]

Answer:

Sabemos que:

MH(x, y) = \frac{2xy}{x+y}

y tenemos que:

MH (a,4) = \frac{2*a*4}{a+4} = \frac{8*a}{a+4} = 6

Con esto podemos encontrar el valor de a:

8*a/(a+ 4) = 6

8*a = 6*(a + 4) = 6*a + 24

8a - 6a = 24

2a = 24

a = 24/2 = 12.

Tambien sabemos que:

MH(8,b) = \frac{2*8*b}{8+b} =\frac{16*b}{8+b} = 12

Y de ahí podemos despejar b:

(16*b)/(b + 8) = 12

16*b = 12*(b + 8) = 12b + 96

16b - 12b = 96

4b = 96

b = 96/4 = 24

Entonces tenemos a = 12 y b = 24, y el MH de a y b es:

MH(12,24) = 2*12*24/(12 + 24) = 24*24/36 = 16

7 0
3 years ago
It is known that the population variance equals 484. With a .95 probability, the sample size that needs to be taken if the desir
Ksju [112]

Answer:

We need a sample size of at least 75.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

\alpha = \frac{1-0.95}{2} = 0.025

Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

So it is z with a pvalue of 1-0.025 = 0.975, so z = 1.96

Now, we find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

The standard deviation is the square root of the variance. So:

\sigma = \sqrt{484} = 22

With a .95 probability, the sample size that needs to be taken if the desired margin of error is 5 or less is

We need a sample size of at least n, in which n is found when M = 5. So

M = z*\frac{\sigma}{\sqrt{n}}

5 = 1.96*\frac{22}{\sqrt{n}}

5\sqrt{n} = 43.12

\sqrt{n} = \frac{43.12}{5}

\sqrt{n} = 8.624

(\sqrt{n})^{2} = (8.624)^{2}

n = 74.4

We need a sample size of at least 75.

6 0
3 years ago
Read 2 more answers
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