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2 years ago

The data set below represents the total number of touchdowns a quarterback threw each season for 10 seasons of play.

Mathematics

2 answers:

tia_tia [17]2 years ago

7 0

The range is 24 touchdowns.

The interquartile range is 8 touchdowns.

Leona [35]2 years ago

5 0

Here are your measures of variability. The range is found by subtracting the highest and the lowest (29-5=24). To find the interquartile range, you will find the median of the lower half of the data and the median of the higher half of sta and subtract these 2 numbers. Here is your list. I have PUT PARENTHESES around the upper and lower quartiles: 5, 17, (18), 20, 20, 21, 23, (26), 28, 29. It is like finding the middle of the entire set of data and then finding the middle of each half. Subtract 26 and 18 to find the interquartile range of 8 touchdowns.

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Emanuel surveyed a random sample of 505050 subscribers to Auto Wheel magazine about the number of cars that they own. Of the sub

Anna [14]

Answer:

The most reasonable estimate for the number of Auto Wheel magazine subscribers who own fewer than 2 is 102

Step-by-step explanation:

We can build an estimate from a sample, and this estimate will be the best estimate for the proportion of the population.

Sample:

50 subscribers

15 own fewer than 2 vehicles

Proportion:

15/50 = 0.3 = 30% own fewer than 2 vehicles.

Based on the data, what is the most reasonable estimate for the number of Auto Wheel magazine subscribers who own fewer than 2 vehicles?

340 subscribers

The most reasonable estimate for this number is 0.3*340.

0.3*340 = 102

The most reasonable estimate for the number of Auto Wheel magazine subscribers who own fewer than 2 is 102

4 0

2 years ago

You know that the current exchange rate for US dollars (USD) to Indian rupees (INR) is 1:45.85, that the exchange rate for India

dem82 [27]

Answer:

Step-by-step explanation:

the answer on edgeunity

3 0

2 years ago

Read 2 more answers

The ratio of money in Brian's wallet to Colin's wallet one day was 5:1. Brian spent £27 that day. Brian now had £3 less than Col

myrzilka [38]

Answer:

Brian had $30 initially.Brian had $30 initially.

Step-by-step explanation:

WE are given the following in the question:

Let Brian have x dollars and Colin have y dollars.

Ration of Brian's money to Colin's money is 5:1. This, we can write the equation:

$\dfrac{x}{y} = \dfrac{5}{1}\\\\x = 5y$

"Brian spent £27 that day. Brian now had £3 less than Colin."

Thus, we can write the equation:

$x-27 = y-3\\x-y = 24$

Solving the two equation by substitution, we get,

$x - y - 24\\5y - y = 24\\4y = 24\\y=6\\x = 5y =30$

Thus, Brian had $30 initially.

7 0

1 year ago

A toddler is allowed to dress himself on Mondays, Wednesdays, and Fridays. For each of his shirt, pants, and shoes, he is equall

avanturin [10]

Answer:

0.0286 = 2.86% probability that today is Monday.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

$P(B|A) = \frac{P(A \cap B)}{P(A)}$

In which

P(B|A) is the probability of event B happening, given that A happened.

$P(A \cap B)$ is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Dressed correctly

Event B: Monday

Probability of being dressed correctly:

100% = 1 out of 4/7(mom dresses).

(0.5)^3 = 0.125 out of 3/7(toddler dresses himself). So

$P(A) = 0.125\frac{3}{7} + \frac{4}{7} = \frac{0.125*3 + 4}{7} = \frac{4.375}{7} = 0.625$

Probability of being dressed correctly and being Monday:

The toddler dresses himself on Monday, so (0.5)^3 = 0.125 probability of him being dressed correctly, 1/7 probability of being Monday, so:

$P(A \cap B) = 0.125\frac{1}{7} = 0.0179$

What is the probability that today is Monday?

$P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.0179}{0.625} = 0.0286$

0.0286 = 2.86% probability that today is Monday.

4 0

2 years ago

Last year, the numbers of skateboards produced per day at a certain factory were normally distributed with a mean of 20,500 skat

BartSMP [9]

Let x be a random variable representing the number of skateboards produced
a.) P(x ≤ 20,555) = P(z ≤ (20,555 - 20,500)/55) = P(z ≤ 1) = 0.84134 = 84.1%

b.) P(x ≥ 20,610) = P(z ≥ (20,610 - 20,500)/55) = P(z ≥ 2) = 1 - P(z < 2) = 1 - 0.97725 = 0.02275 = 2.3%

c.) P(x ≤ 20,445) = P(z ≤ (20,445 - 20,500)/55) = P(z ≤ -1) = 1 - P(z ≤ 1) = 1 - 0.84134 = 0.15866 = 15.9%

5 0

2 years ago