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3 years ago

How do you do this problem ?

Mathematics

1 answer:

miskamm [114]3 years ago

6 0

You could do all of the work with the multiplying, but what I think it wants you to see is that you can divide like numbers by subtracting exponents.

so if we did that we would get $(-2^{2})(-3^3)(-5^1)$

this in turn would give us 4 x -9 x -5 = 180

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Pls help me now!! (T^T)

Minchanka [31]

Vas happenin!!

1. Since it’s subtracting, another word for subtracting is difference

B is correct

2. There’s both a sum and difference

C is correct

Hope this helps

-Zayn Malik

6 0

2 years ago

5 / 1/6 + 1/x+1 Math help please

zubka84 [21]

The common denomenators of the denomenaotrs is 6(x+1)
times both the fraction by (6(x+1))/(6(x+1))
we get
(5(6)(x+1))/((x+1)+6)=(30(x+1))/(x+7)=(30x+30)/(x+7)

5 0

3 years ago

Read 2 more answers

The value of (-8/13×26/-3)

ahrayia [7]

Answer:

Step-by-step explanation:

Given the expression -8/13×26/-3, getting the equivalent expression means multiplying the two fractions -8/13 and 26/-3 together as shown;

$\frac{-8}{13} * \frac{26}{-3}\\ \\= \frac{-8*26}{13*-3}\\ \\multiplying\ out \ the\ numerator\ and\ denominator;\\\\= \frac{-208}{-39} \\\\cancelling out the negative sign we will have;\\\\= \frac{208}{39}\\\\= 5.33\\The\ resulting\ value\ is\ 5.33$

3 0

3 years ago

Suppose that in a random selection of 100 colored candies, 24% of them are blue. The candy company claims that the percentage

LuckyWell [14K]

Answer:

We conclude that the percentage of blue candies is equal to 23% at 0.01 significance level.

Step-by-step explanation:

We are given that in a random selection of 100 colored candies, 24% of them are blue.

The candy company claims that the percentage of blue candies is equal to 23%.

Let p = percentage of blue candies.

So, Null Hypothesis, $H_0$ : p = 23% {means that the percentage of blue candies is equal to 23%}

Alternate Hypothesis, $H_A$ : p $\neq$ 23% {means that the percentage of blue candies is different from 23%}

The test statistics that would be used here One-sample z proportion test statistics;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n}} }$ ~ N(0,1)

where, $\hat p$ = sample proportion of blue colored candies = 24%

n = sample of colored candies = 100

So, test statistics = $\frac{0.24-0.23}{\sqrt{\frac{0.24(1-0.24)}{100}} }$

= 0.234

The value of z test statistics is 0.234.

Now, at 0.01 significance level the z table gives critical values of -2.5758 and 2.5758 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that the percentage of blue candies is equal to 23%.

6 0

3 years ago

Please help me. It’s urgent.

sergeinik [125]

The answer to the question

5 0

3 years ago