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raketka [301]
3 years ago
14

ASAP there are three marbles in a bag. One is red and two are black. What is the probability of picking a black marble first, pu

tting it back in the bag and then picking a black marble? Use the following probability to find the answer.

Mathematics
2 answers:
FrozenT [24]3 years ago
6 0

Answer:

\frac{4}{9}

Step-by-step explanation:

p =  \frac{favorable \: outcomes}{total \: outcomes}  =  \frac{4}{9}

Marina86 [1]3 years ago
6 0
<h3>Answer:  Choice D)  4/9</h3>

=============================================================

Explanation:

The probability you get a black marble on the first selection is 2/3 since we have 2 black marbles out of 2+1 = 3 total.

We put the marble back and then we have 2/3 as the probability of selecting another black marble on the second try. Nothing has changed because we put the marble back. That means the events are independent.

So we get (2/3)*(2/3) = 4/9 as the probability of selecting 2 black marbles in a row (with replacement).

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12.5*<img src="https://tex.z-dn.net/?f=12.5%20%2A17.9%3D" id="TexFormula1" title="12.5 *17.9=" alt="12.5 *17.9=" align="absmiddl
kkurt [141]
The answer is 223.75 because you multiply 12.5 and 17.9
5 0
3 years ago
lori says that 8 is not a factor of 80 because 8 does not end in zero. Does Lori's statement make sense? Explain
77julia77 [94]
You have to know that the zero has no value inless its in a place value .the 8 is a factor of 80 because 80 is 10 times as much as 8
5 0
3 years ago
Read 2 more answers
Given sin x =-4/5 and x is in quadrent 3, what is the value of tan x/2
4vir4ik [10]

Answer:

We can write sin x in terms of tan x/2 using the formula:

⇒ sin x = (2 tan (x/2)) / (1 + tan2(x/2))

Therefore, using the above formula, we can find the values of tan x/2 by putting the value of sin x.

⇒ -4/5 = (2 tan (x/2)) / (1 + tan2(x/2))

Now, if we replace tan (x/2) by y, we get a quadratic equation:

⇒ 0.8y2 + 2y + 0.8 = 0

⇒ 2y2 + 5y + 2 = 0

By using the quadratic formula, we get y = -0.5, -2

Hence, the value of tan (x/2) = -0.5, -2

Now, we have two solutions of tan (x/2).

Now, let's check for the ideal solution using the formula tan x = (2 tan (x/2)) / (1 - tan2(x/2)).

For tan (x/2) = -0.5:

⇒ tan x = 2(-0.5) / 1 - (-0.5)2 = -4/3

It is also given that x lies in the third quadrant. We know that tan is positive in the third quadrant, and here we get tan x = -4/3 which is negative.

Hence, we can say that tan (x/2) = -0.5 is not a correct solution. Hence it is rejected.

Now let's check for tan (x/2) = -2.

⇒ tan x = 2(-2) / 1 - (-2)2 = 4/3

Here, we get tan x = 4/3 which is positive.

Hence, we can say that tan (x/2) = -2 is a correct solution.

5 0
3 years ago
REPOST*
xeze [42]

#1

  • -6-3-5

Hence inequality will be

\\ \sf\longmapsto -6\leqslant 5

#2.

  • 3<5

But we need inequality

\\ \sf\longmapsto 3\leqslant 5

#3

The inequality given by

\\ \sf\longmapsto -6\leqslant 3\leqslant 5

7 0
2 years ago
Read 2 more answers
A professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their class
KIM [24]

Answer:

We conclude that seniors skip more than 2% of their classes at 0.01 level of significance.

Step-by-step explanation:

We are given that a professor wishes to discover if seniors skip more classes than freshmen. Suppose he knows that freshmen skip 2% of their classes.

He randomly samples a group of seniors and out of 2521 classes, the group skipped 77.

<u><em /></u>

<u><em>Let p = percentage of seniors who skip their classes.</em></u>

So, Null Hypothesis, H_0 : p \leq 2%   {means that seniors skip less than or equal to 2% of their classes}

Alternate Hypothesis, H_A : p > 2%   {means that seniors skip more than 2% of their classes}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

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

where, \hat p = sample proportion of seniors who skipped their classes = \frac{77}{2521}

           n = sample of classes = 2521

So, <u><em>test statistics</em></u>  =  \frac{\frac{77}{2521} -0.02}{{\sqrt{\frac{\frac{77}{2521}(1-\frac{77}{2521})}{2521} } } } }

                               =  3.08

The value of the test statistics is 3.08.

Now at 0.01 significance level, <u>the z table gives critical value of 2.3263 for right-tailed test</u>. Since our test statistics is more than the critical value of z as 2.3263 < 3.08, 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 seniors skip more than 2% of their classes.

6 0
3 years ago
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