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astraxan [27]
2 years ago
6

Help me solve this please

Mathematics
1 answer:
Anvisha [2.4K]2 years ago
8 0

Answer:

2/5-1/10 = 3/10

1/5+7/10 = 9/10

Step-by-step explanation:

4/10-1/10 = 3/10

2/10+7/10 = 9/10

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If is an angle bisector of ∠QOR and ∠QOP = 71°, then find the angle measure of ∠QOR. Question 8 options: A) 71° B) 142° C) 35.5°
zaharov [31]

Answer:  B. 142°.

Step-by-step explanation:

Given: OP is an angle bisector of ∠QOR and ∠QOP = 71°.

We know that an angle bisector divides an angle into two equal parts.

So, if OP is an angle bisector of ∠QOR and ∠QOP = 71°.

Then, the angle measure of ∠QOR = Twice of ∠QOP

⇒ Angle measure of ∠QOR = 2 (71°)

⇒ Angle measure of ∠QOR = 142°

Hence, correct option is B. 142°.

8 0
3 years ago
Suppose that past history shows that 60% of college students prefer brand c cola. a sample of 5 students is to be selected. base
creativ13 [48]
The probability is 0.337.

The probability that more than 3 prefer brand c is the same as finding the probability that 4 or 5 prefer brand c.  Using a binomial distribution,

_5C_4(0.6)^4(0.4)^{5-4}+_5C_5(0.6)^5(0.4)^{5-5}
\\
\\=\frac{5!}{4!1!}(0.6)^4(0.4)^1+\frac{5!}{5!0!}(0.6)^5(0.4)^0
\\
\\=5(0.6)^4(0.4)+1(0.6)^5(1)=0.33696\approx0.337
4 0
3 years ago
These box plots show the prices for two different brands of shoes.
nignag [31]

Answer: Choice C

The interquartile range (IQR) for brand A, $10, is less than the IQR for brand B, $20.

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

Explanation:

Let's go through the answer choices one by one to see which is true and which is false. Also, we'll see which is helpful when it comes to the spread.

----------------------

Choice A

The IQR is found by subtracting the values of Q1 and Q3, which are the left and right edges of the box in that order.

For boxplot A, we have an IQR = Q3-Q1 = 80-70 = 10.

So we can immediately rule out choice A because the IQR for boxplot A is not 20, but instead it's 10.

----------------------

Choice B

The statement made in choice B is true, since the center vertical lines represent the median, but it's not useful to determine the spread. The median only tells us the center of the distribution. It doesn't tell us how spread out the distribution is.

Imagine you didn't have access to the graph. Now if you're told that the median for brand A is larger than brand B's median, then you'd probably think boxplot A is more spread out; however, the boxplot diagram shows the opposite is the case.

So again the median isn't used here. We can rule out choice B.

----------------------

Choice C

This statement is true. We subtract the box edge values to get the IQR values for each boxplot

For brand A we have IQR = Q3-Q1 = 80-70 = 10

For brand B we have IQR = Q3-Q1 = 70-50 = 20

This is the answer we're after.

The IQR is one tool we can use to measure the spread. The range is another option. We could also use the variance or the standard deviation (two slightly similar ideas) to measure the spread.

As you can see from the diagram, the larger the IQR, the more spread out the data will be.

----------------------

Choice D

This is false because earlier we've shown 10 is the IQR for brand A.

7 0
3 years ago
You can pack 6 cans in 1 box. Write a proportion to find the number of
seraphim [82]

Answer:

Step-by-step explanation:

You can fit 6 cans in a box. So now there are 12 boxes. Therefore you have to multiply 6 times 12 which should give you 72. You can fit 72 cans in 12 boxes total

6 0
2 years ago
Nine out of ten students prefer math class over lunch. How many students do not prefer math if 200 students were asked?
fredd [130]
So, 9 out of 10 students prefer class, so, obviously, it would not be 100. It would be 150. So in conclusion, 50 students prefer lunch over math class, whilst 150 other students prefer math class over lunch.
3 0
3 years ago
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