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S_A_V [24]
3 years ago
14

A box contains 6 red marbles, 4 green marbles, 3 blue marbles, and 2 yellow marbles. If Greg chooses one marble at random 300 ti

mes, how many times would he expect the marble to be blue?
Mathematics
1 answer:
dusya [7]3 years ago
3 0

Answer:

60 times

Step-by-step explanation:

There are 6+4+3+2=15 marbles in total. Out of those, 3 are blue, so the probability of pulling a blue marble is \frac{3}{15}=\frac{1}{5}. However, this is only the probability of pulling a blue marble when 1 marble is pulled at random. Therefore, we can expect a blue marble to be pulled \frac{1}{5}*300=60 times if 1 marble is pulled 300 times at random. Hope this helps!

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Factorise = <br> 5m + 20
taurus [48]
So 20=2 itmes m2 times 5

5m=5 times m

so
5m=20
5 times m=2 times 2 times 5
divid eboth sides y 5
m=2 times 2
m=4
8 0
4 years ago
The home run percentage is the number of home runs per 100 times at bat. A random sample of 43 professional baseball players gav
Andru [333]

Step-by-step explanation:

(a) Yes, if you enter all 43 values into your calculator, you calculator should report:

xbar = 2.293

s = 1.401

(b)

Note: Most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 1.684 * 1.401 / sqrt(43) = 1.933

2.293 - 1.684 * 1.401 / sqrt(43) = 2.653

Answer: (1.933, 2.653)

Note: To find the t-value that allows us to be 90% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.90)/2 = .05 or up from 90% depending on your t-table. So, the t-critical value is 1.684.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 1.681).

2.293 +/- 1.681 * 1.401 / sqrt(43)

(1.934, 2.652)

Note: Some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 90% CI is:

2.293 +/- 1.645 * 1.401 / sqrt(43)

(1.942, 2.644)

Note: To find the z-value that allows us to be 90% confident, (1) using the z-table, look up (1-.90)/2 = .05 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .05 or up from 90% depending on your t-table. Either way, the z-critical value is 1.645.

(c)

Note: Again, most professors say that is sigma = the population standard deviation is unknown (as it is unknown here), you should construct a t-confidence interval.

xbar +/- t * s / sqrt(n)

2.293 - 2.704 * 1.401 / sqrt(43) = 1.715

2.293 - 2.704 * 1.401 / sqrt(43) = 2.871

Answer: (1.715, 2.871)

Note: To find the t-value that allows us to be 99% confident, go across from df = 43-1 = 42 (round down to 40 to be conservative since 42 in not in the table) and down from (1-.99)/2 = .005 or up from 99% depending on your t-table. So, the t-critical value is 2.704.

Note: If you can use the TI-83/84, it will construct the following CI using df = 42 (ie t = 2.698).

2.293 +/- 2.698 * 1.401 / sqrt(43)

(1.717, 2.869)

Note: Again, some professors want you to construct a z-CI when the sample size is large. If your professor says this, the correct 99% CI is:

2.293 +/- 2.576 * 1.401 / sqrt(43)

(1.742, 2.843)

Note: To find the z-value that allows us to be 99% confident, (1) using the z-table, look up (1-.99)/2 = .005 inside the z-table, or (2) using the t-table, go across from infinity df (= z-values) and down from .005 or up from 99% depending on your t-table. Either way, the z-critical value is 2.576.

(d)

Tim Huelett 2.5

Since 2.5 falls between (1.715, 2.871), we see that Tim Huelett falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Herb Hunter 2.0

Since 2.0 falls between (1.715, 2.871), we see that Herb Hunter falls in the 99% CI range. So, his home run percentage is NOT significantly different than the population average.

Jackie Jensen 3.8.

Since 3.8 falls above (1.715, 2.871), we see that Jackie Jensen falls in the 99% CI range. So, his home run percentage IS significantly GREATER than the population average.

(e)

Because of the Central Limit Theorem (CLT), since our sample size is large, we do NOT have to make the normality assumption since the CLT tells us that the sampling distribution of xbar will be approximatley normal even if the underlying population distribution is not.

6 0
3 years ago
Question 2 of 10<br> Which describes the graph of y = (x-4)2 - 1?
WITCHER [35]

y=(x-4)^2-1:   y=-\frac{4}{5}

Sure hope this helps you

y = -4/5

4 0
2 years ago
Select the correct answer.
Crazy boy [7]

check the picture below.

now, if we split that triangle like so with that red line, we can use the right-triangle there, to get θ.


\bf tan(\theta )=\cfrac{\stackrel{opposite}{3}}{\stackrel{adjacent}{3}}\implies tan(\theta )=1\implies tan^{-1}(\theta )=tan^{-1}(1)\implies \measuredangle \theta =\cfrac{\pi }{4}

8 0
3 years ago
19/24 as a decimal rounded to the nearest hundreths
Evgen [1.6K]

Answer:

0.79

Step-by-step explanation:

19 / 24 is 0.79166667

So it's 0.79

8 0
3 years ago
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