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nataly862011 [7]
4 years ago
12

Someone help me with this question ​

Mathematics
2 answers:
zalisa [80]4 years ago
7 0

Answer:

Step-by-step explanation:

a) Complete the table

                          Monday             Tuesday            Wednesday             Total

Female       21-14=7                 38-13-7= 18                 13                          38

Male              14                       33-18= 15                 26-13 =13            80-38=42

Total        80-33-26= 21                 33                          26                      80

b) P(the student is a female) = 38/80= 0.475

c) P(the student visited the library on Tuesday) = 33/80= 0.4125

leonid [27]4 years ago
5 0

Answer:

a) Monday Total: 21.

Monday Female: 7.

Tuesday Female: 18.

Tuesday Male: 15

Wednesday Male: 13

Male Total: 42.

b) 47.5% probability that the student is a female.

Step-by-step explanation:

<u>a)</u> How to find Monday Total: We have the totals for Tuesday and Wednesday, and we know the total of all three days is 80. So, we need to find the difference between Tuesday + Wednesday & all 3 days, to find the total of Monday. What we do is add 33 + 26 which equals to 59. Then, we remove 59 from 80 (80-59). The answer we get is 21, which is the difference between the two. That's Monday's total.

How to find Male Wednesday: We already have Wednesday Female, and the total for Wednesday. But now we're finding the Wednesday Male. Find the difference between Wednesday Total and Female (26-13). The answer is 13! The Wednesday Male is therefore 13.

How to find Male Total: We have to find the difference between the complete total (80) and the Female total (38). Subtract: (80-38). The answer you get is 42!

How to find Tuesday Male: Now that we have the Wednesday Male, and the Male Total, we can easily find the Tuesday Male! The Male Total is 42, and we know the Male Monday is 14, along with the Male Wednesday being  13. Add Male Monday and Male Wednesday together. (13+14=27). Now take that and remove it from 42. (42-27). The answer we get is 15. 15 is our Tuesday Male.

How to find Monday Female: We already know what the Monday Total is from previous calculations(21), so all we have to do now is find the difference from the Monday Total and the Male Monday to find the Female Monday. Subtract: (21-14) The answer we will get is 7. The Monday female is 7.

How to find Tuesday Female: We already have Tuesday Male, and the Tuesday Total, so we subtract the Tuesday Male from the Tuesday Total. (33-15). The answer we get is 18.

<u>b)</u> The answer would be 47.5% because 38 is 47.5% of 80. How? Divide 38 by 80. The answer would be 0.475. Now, multiply that answer by 100. You get 47.5! That's the percentage.

This is my first question that I've answered, I'm sorry if there are any mistakes!

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Organizers of an outdoor summer concert in Toronto are concerned about the weather conditions on the day of the concert. They wi
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Answer:

The value of the expected profit from the concert is 8,910

Step-by-step explanation:

Profit on a clear day X₁ = 36,000 with  13% probability.  

i.e  X₁ = 36000

    P(X₁) = 0.13

Profit on a cloudy day = 17,000 with 39% probability.

i.e  X₂ = 17000

    P(X₂) = 0.39

else,

loss of 5,000 if it rains with the probability of 48%.

i.e  X₃ = 5000

    P(X₃) = 0.48

The value of the expected profit from the concert is obtained as follows

Expected Value = (36,000*0.13) + (17,000*0.39) - (5,000*0.48)

                           = 4,680 + 6,630 - 2,400

                           = 8,910

8 0
3 years ago
In a random sample of 80 teenagers, the average number of texts handled in a day is 50. The 96% confidence interval for the mean
Nastasia [14]

Answer:

a) \bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

b) ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

\bar X represent the sample mean  

\mu population mean (variable of interest)  

\sigma represent the population standard deviation  

n=80 represent the sample size  

Solution to the problem

Part a

The confidence interval for the mean is given by the following formula:  

\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}} (1)  

For this case we can calculate the mean like this:

\bar X =\frac{46+54}{2}=50

And the margin of error is given by:

ME= \frac{54-46}{2}= 4

The confidence level is 0.96 and the significance level is \alpha=1-0.96=0.04 and the value of \alpha/2 =0.02 and the margin of error is given by:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}

We can calculate the critical value and we got:

z_{\alpha/2} = 2.05

And if we solve for the deviation like this:

\sigma = ME * \frac{\sqrt{n}}{z_{\alpha/2}}

And replacing we got:

\sigma =4 *\frac{\sqrt{80}}{2.05} =17.45

Part b

For this case is the sample size is doubled the margin of error would be:

ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}=2.05 *\frac{17.45}{\sqrt{160}}=2.828

And as we can see that the margin of error would be lower than the original value of 4, the margin of error would be reduced by a factor \sqrt{2}

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3 years ago
A and B are two similar shapes the area of shape A is 200cm2 calculate the area of shape B
barxatty [35]

Answer:

Area of shape B = 312.5 cm²

Step-by-step explanation:

Ratio of Area of A : Area of B = 12² : 15²

Given that area of A is 200 cm², let the area of shape B = x

Therefore:

200:x = 12²:15²

200/x = 144/225

Cross multiply

144x = 200*225

144x = 45,000

x = 45,000/144

x = 312.5 cm²

Area of shape B = 312.5 cm²

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Para decorar una pared se disponen de tiras de papel azules de 15 cm, verdes de 20 cm, y rojas de 25 cm. En la pared se quiere a
den301095 [7]

Answer:

a) Smallest line that can be made with each color = 300 cm

b) Total strips should be used = 47 strips

c) Total strips used of blue color = 20

  Total strips used of green color = 15

  Total strips used of red color = 12

Step-by-step explanation:

Given - To decorate a wall, there are 15 cm blue, 20 cm green, and 25 cm red strips of paper. On the wall you want to build three lines of the same size, one of each color and without cutting any strip.

To find - a) How long is the smallest line that can be made with each color?

              b) How many strips should be used?

              c) How many of each color?

Proof -

a)

For the smallest line that can be made with each color, we just have to find the lcm (least common multiple) of the 3 srtips.

Firstly,

Decompose the 3 strips to its prime factors , we get

15 = 3×5

20 = 2²×5

25 = 5²

So,

The Lcm(15, 20, 25) = 3×2²×5² = 3×4×25 = 300

∴ we get

Smallest line that can be made with each color = 300 cm

b)

Now,

Total strips used = 300 cm

Strips used by 15 cm blue = \frac{300}{15} = 20 strips

Strips used by 20 cm green = \frac{300}{20} = 15 strips

Strips used by 25 cm red = \frac{300}{25} = 12 strips

So,

Total strips should be used = 20 + 15 + 12 = 47 strips

c)

Total strips used of blue color = 20

Total strips used of green color = 15

Total strips used of red color = 12

6 0
3 years ago
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