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PilotLPTM [1.2K]
3 years ago
15

How do I know if two ratios are equivalent

Mathematics
2 answers:
Luba_88 [7]3 years ago
6 0

Answer:

Step-by-step explanation: there are 24

jok3333 [9.3K]3 years ago
3 0

Answer:

So to have multiple ratios, you may want to determine whether they are equal or if one of them is larger.

To compare ratios, you need to have a common second number.

By multiplying each ratio by the second number of the other ratio, you can determine if they are equivalent.

have a good day and be safe

-HOPS

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One hundred 50 cent coins have a mass of 690g. What is the mass of one 50 cent coin?
olga nikolaevna [1]
You divide 690 by 100 which is 6.9
Therefore, one 50cent coin is 6.9g
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owl monkey sleep during the day waking about 15minutes after sundown to find food.At midnight,they rest for an hour or two,then
Ket [755]
You do 27x27 and get 729 is going to be your answer
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If 2a-b:2a+b=1:2find the value of a:b​
weqwewe [10]

Answer & Step-by-step explanation:

2a - b : 2a + b = 1 : 2

2a - b = 1

-                         Elimination

2a + b = 2

-b - (b) = -1

-2b = -1

b = 0.5

---------------------------------------------------------------------------------

2a + b = 2           Substitution

2a + 0.5 = 2

2a = 1.5

a = 0.75

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a:b

0.75 : 0.5

or

3/4 : 1/2

Hope this helps!!!

 

3 0
3 years ago
Y = –3x 6 y = 9 what is the solution to the system of equations? (–21, 9) (9, –21) (–1, 9) (9, –1)
Romashka [77]

we have

<u>The system of equations</u>

y=-3x+6 -------> equation 1

y=9 -------> equation 2

Substitute equation 2 in equation 1

9=-3x+6

subtract 6 both sides

9-6=-3x+6-6

3=-3x

Divide by -3 both sides

x=-1

the solution is the point (-1,9)

therefore

the answer is

the solution to the system of equations is the point (-1,9)


4 0
3 years ago
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The weight of an adult swan is normally distributed with a mean of 26 pounds and a standard deviation of 7.2 pounds. A farmer ra
Snezhnost [94]
Let X denote the random variable for the weight of a swan. Then each swan in the sample of 36 selected by the farmer can be assigned a weight denoted by X_1,\ldots,X_{36}, each independently and identically distributed with distribution X_i\sim\mathcal N(26,7.2).

You want to find

\mathbb P(X_1+\cdots+X_{36}>1000)=\mathbb P\left(\displaystyle\sum_{i=1}^{36}X_i>1000\right)

Note that the left side is 36 times the average of the weights of the swans in the sample, i.e. the probability above is equivalent to

\mathbb P\left(36\displaystyle\sum_{i=1}^{36}\frac{X_i}{36}>1000\right)=\mathbb P\left(\overline X>\dfrac{1000}{36}\right)

Recall that if X\sim\mathcal N(\mu,\sigma), then the sampling distribution \overline X=\displaystyle\sum_{i=1}^n\frac{X_i}n\sim\mathcal N\left(\mu,\dfrac\sigma{\sqrt n}\right) with n being the size of the sample.

Transforming to the standard normal distribution, you have

Z=\dfrac{\overline X-\mu_{\overline X}}{\sigma_{\overline X}}=\sqrt n\dfrac{\overline X-\mu}{\sigma}

so that in this case,

Z=6\dfrac{\overline X-26}{7.2}

and the probability is equivalent to

\mathbb P\left(\overline X>\dfrac{1000}{36}\right)=\mathbb P\left(6\dfrac{\overline X-26}{7.2}>6\dfrac{\frac{1000}{36}-26}{7.2}\right)
=\mathbb P(Z>1.481)\approx0.0693
5 0
3 years ago
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