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marissa [1.9K]
3 years ago
8

When 6a^2 b^3• -3a^5 b^3 multiplied, the product of the coefficients is .

Mathematics
2 answers:
ad-work [718]3 years ago
7 0

Answer: last option

Step-by-step explanation:

To solve this problem you must apply the properties of the exponents. WHen you have two powers with equal base, you must add the exponents.

Then you have:

(6a^{2}b^{3})(-3a^{5}b^{3})=-18a^{(2+5)}b^{(3+3)}=-18a^{7}b^{6}

Therefore as you can see, the coefficient is -18

Vlad1618 [11]3 years ago
5 0

Answer:

-18

Step-by-step explanation:

Question says to multiply 6a^2 b^3 and -3a^5 b^3.

to find the product of the coefficients. And select correct choice from given list:

3

-3

18

-18

Coefficient of 6a^2 b^3 is 6

Coefficient of -3a^5 b^3 is -3

Then product of coefficients = (6)(-3)=-18

Hence -18 is correct choice.

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Find an equation of the line through (3,6) and parallel to y = 4x - 7.
FromTheMoon [43]

Answer:

y=4x-6

Step-by-step explanation:

Okay, to do this we must understand what parallel means.

parallel means that that one line will NEVER touch the other. In order for this to happen they must be increasing (or decreasing) by the SAME rate. I.e the slope of your line must be the same as the lin in question thus our equation looks like,

y=4x+c.

Now lets find c.

To do this, plug the point (3,6) into the equation

6=4(3)+C

6=12+C

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Thus now that we know C, our equaiton is

y=4x-6

4 0
3 years ago
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A public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes. Kar
ch4aika [34]

Answer:

We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let \mu = <u><em>mean waiting time for bus number 14.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 minutes      {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, H_A : \mu < 10 minutes    {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean waiting time = 7.8 minutes

             s = sample standard deviation = 2.5 minutes

             n = sample of different occasions = 18

So, <u><em>test statistics</em></u> =  \frac{7.8-10}{\frac{2.5}{\sqrt{18} } }  ~ t_1_7

                              =  -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

5 0
4 years ago
F(x) = -x2 – 9.2 – 4
drek231 [11]

Answer:

what is the question?

Step-by-step explanation:

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3 years ago
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Quadrilateral ABCD is similiar to quadrilateral EFGH. The lengths of the three longest sides in quadrilateral ABCD are 20 feet,
Harman [31]

First, let's list the lengths of the sides in descending order.

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We are looking for c which corresponds to 18.

14 is to 6 is as 18 is to c

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7c = 3 * 18

7c = 54

c = 54/7 = 7 5/7

Answer: 7 5/7 feet

5 0
4 years ago
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