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Roman55 [17]
3 years ago
8

Simon draws a rectangle with two sides that are 2 inches long and two sides that are 3 inches long.Chaz draws a figure with the

same measurements,but it is not a rectangle.What figure could it be.
Mathematics
1 answer:
7nadin3 [17]3 years ago
3 0
The other shape might be a trapezoid.
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Miguel is playing a game in which a box contains four chips with numbers written on them. Two of the chips have the number 1, on
insens350 [35]
1) We have that there are in total 6 outcomes If we name the chips by 1a, 1b, 3 ,5 the combinations are: 1a,3 \ 1b, 3 \1a, 5\ 1b, 5\ 3,5\1a,1b. Of those outcomes, only one give Miguel a profit, 1-1. THen he gets 2 dollars and in the other cases he lose 1 dollar. Thus, there is a 1/6 probability that he gets 2$ and a 5/6 probability that he loses 1$.
2) We can calculate the expected value of the game with the following: E=\frac{1}{6}*2- \frac{5}{6} *1. In general, the formula is E= \sum{p*V} where E is the expected value, p the probability of each event and V the value of each event. This gives a result of E=2/6-5/6=3/6=0.5$ Hence, Miguel loses half a dollar ever y time he plays.
3) We can adjust the value v of the winning event and since we want to have a fair game, the expecation at the end must be 0 (he must neither win or lose on average). Thus, we need to solve the equation for v:
0=\frac{1}{6}v -\frac{5}{6} =0. Multiplying by 6 both parts, we get v-5=0 or that v=5$. Hence, we must give 5$ if 1-1 happens, not 2.
4) So, we have that the probability that you get a red or purple or yellow sector is 2/7. We have that the probability for the blue sector is only 1/7 since there are 7 vectors and only one is blue. Similarly, the 2nd row of the table needs to be filled with the product of probability and expectations. Hence, for the red sector we have 2/7*(-1)=-2/7, for the yellow sector we have 2/7*1=2/7, for the purple sector it is 2/7*0=0, for the blue sector 1/7*3=3/7. The average payoff is given by adding all these, hence it is 3/7.
5) We can approach the problem just like above and set up an equation the value of one sector as an unknown. But here, we can be smarter and notice that the average outcome is equal to the average outcome of the blue sector.  Hence, we can get a fair game if we make the value of the blue sector 0. If this is the case, the sum of the other sectors is 0 too (-2/7+0+2/7) and the expected value is also zero.
6) We want to maximize the points that he is getting. If he shoots three points, he will get 3 points with a probability of 0.30. Hence the average payoff is 0.30*3=0.90. If he passes to his teammate, he will shoot for 2 points, with a success rate of 0.48. Hence, the average payoff is 0.48*2=0.96. We see that he should pass to his player since 0.06 more points are scored on average.
7) Let us use the expections formula we mentioned in 1. Substituting the possibilities and the values for all 4 events (each event is the different profit of the business at the end of the year).
E=0.2*(-10000)+0.4*0+0.3*5000+0.1*8000=-2000+0+1500+800=300$
This is the average payoff of the firm per year.
8) The firm goes even when the total profits equal the investment. Suppose we have that the firm has x years in business. Then x*300=1200 must be satisfied, since the investment is 1200$ and the payoff per year is 300$. We get that x=4. Hence, Claire will get her investment back in 4 years.
8 0
3 years ago
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What is the base of a triangle that has a height of 6 centimeters and an area of 18 centimeters? Use the formula , where A repre
harina [27]

Answer:

The base of the triangle is 6cm.

Step-by-step explanation:

The find the base of a triangle, use this formula.

Formula : 1/2 x b x h

<em>If you would like some more explanation, see the attached image.</em>

5 0
3 years ago
Marcie jogs each morning. The relationship between the total number of miles she jogged this week, m, to the total number of mil
Phantasy [73]

Answer: Marcie jogged twice as many miles this week as she jogged last week

Step-by-step explanation:

5 0
3 years ago
DRAW AN AREA MODEL WITH THIS EQUATION PEOPLE LITERALLY JUST SOLVE THE MULTIPICATION ​
Iteru [2.4K]

Answer:

You need four squares.

On the top left square, you put the number 20 over it, and the number 30 to its left. Put the number 600 in the square.

Next, the bottom left square. But the number 4 on the left side of the square, and put the number 80 in the square.

Now, the top right box. put the number 8 over it, and fill it with the number 240.

Last, fill the bottom right square with the number 32.

7 0
3 years ago
Read 2 more answers
The scores of students on the ACT college entrance exam in a recent year had the normal distribution with mean  =18.6 and stand
Maurinko [17]

Answer:

a) 33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) 0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, the sample means with size n of at least 30 can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 18.6, \sigma = 5.9

a) What is the probability that a single student randomly chosen from all those taking the test scores 21 or higher?

This is 1 subtracted by the pvalue of Z when X = 21. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{21 - 18.6}{5.4}

Z = 0.44

Z = 0.44 has a pvalue of 0.67

1 - 0.67 = 0.33

33% probability that a single student randomly chosen from all those taking the test scores 21 or higher.

b) The average score of the 76 students at Northside High who took the test was x =20.4. What is the probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher?

Now we have n = 76, s = \frac{5.9}{\sqrt{76}} = 0.6768

This probability is 1 subtracted by the pvalue of Z when X = 20.4. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{20.4 - 18.6}{0.6768}

Z = 2.66

Z = 2.66 has a pvalue of 0.9961

1 - 0.9961 = 0.0039

0.39% probability that the mean score for 76 students randomly selected from all who took the test nationally is 20.4 or higher

4 0
3 years ago
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