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Softa [21]
1 year ago
10

Hector is competing in a 42 mile bicycle race. He has already completed 18 miles of the race and is traveling at a constant spee

d of 12 miles per hour when Wanda starts the race. Wanda is traveling at a constant speed of 16 miles per hour. Write and solve an equation to find when Wanda will catch up to Hector. Will Wanda catch up to Hector before the race is complete? Explain. At what constant speed would Wanda travel to catch up with Hector at the finish line? Explain.

Mathematics
1 answer:
Lisa [10]1 year ago
7 0
Wanda reaches hector at 4.5h. Wanda can not reach hector before the race ends, since at the speed he is going (16m / h) he would reach when both are at mile 72 and the race is 42 miles.
Wanda needs to increase her speed to 21m / h to catch Hector at the finish line.I attach the answers.

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Alana and two of her friends evenly split the cost of a meal. They determine that each person owes $14.68. How much was the enti
AlekseyPX

Answer:

44.04 bucks

Step-by-step explanation:

14.68 times 3 because 1 plus 2 is 3 friends. And 14.68 is how much they EACH paid.

6 0
1 year ago
Read 2 more answers
A normal distribution curve, where x = 70 and σ = 15, was created by a teacher using her students’ grades. What information abou
mash [69]

Answer:

The median and mode of the students grade is 70.

Most of the students scored between 40 and 100.

Step-by-step explanation:

From the provided information it can be seen that the mean of the distribution is, <em>μ</em> = 70 and the standard deviation is, <em>σ</em> = 15.

For a Normal distributed data the mean, median and mode are the same.

So, the median and mode of the students grade is 70.

The standard deviation of the data represents the spread of the observation, i.e. how dispersed the values are along the curve.

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is a shortcut used to recall that 68.27%, 95.45% and 99.73% of the values of a Normally distributed data lie within one, two and three standard deviations of the mean, respectively.

P(\mu-\sigma  

P(\mu-2\sigma

P(\mu-3\sigma

Assuming that maximum marks of the exam is 100, it can be said that most of the students scored between 40 and 100.

3 0
1 year ago
Read 2 more answers
Consider a single spin of the spinner.
alex41 [277]

Answer:

"landing on a shaded portion and landing on a 3"

"landing on an unshaded portion and landing on a number less than 2 "

Step-by-step explanation:

Mutually exclusive means the events will have no intersection.

Let's look at your first choice:

"landing on a shaded portion and landing on an even number"

Landing on a shaded portion would be 1 or 4.

Landing on an even number would be 2 or 4.

There is an intersection (they contain a common element), the 4.

These events are not mutually exclusive.

Let's look at your second choice:

"landing on a shaded portion and landing on a number greater than 3"

Landing on a shaded portion would be 1 or 4.

Landing on a number greater than 3 would be just 4.

There is an intersection; they both contain 4.

These events are not mutually exclusive.

Let's look at your third choice:

"landing on a shaded portion and landing on a 3"

Landing on a shaded portion would be 1 or 4.

Landing on 3 would just be 3.

There is no common elements in the lists listed.  These events have no intersection.

These events are mutually exclusive.

Let's look at your fourth choice:

"landing on an unshaded portion and landing on an odd number"

Landing on a unshaded portion would be 2 or 3.

Landing on an odd number would be 1 or 3.

There is an intersection; they both have 3 in common.

These events are not mutually exclusive.

Let's look at your fifth choice:

"landing on an unshaded portion and landing on a number less than 2 "

Landing on an unshaded portion would 2 or 3.

Landing on a number less than 2 would be 1.

There is no intersection.

These events are mutually exclusive.

6 0
1 year ago
Read 2 more answers
Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar(s).
Tanzania [10]

Answer:

Approximately, 159 men weighs more than 165 pounds and  159 men weighs less than 135 pounds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 150 pounds

Standard Deviation, σ = 15

We are given that the distribution of weights of 1000 men is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P( men weighing more than 165 pounds)

P(x > 165)

P( x > 165) = P( z > \displaystyle\frac{165 - 150}{15}) = P(z > 1)

= 1 - P(z \leq 1)

Calculation the value from standard normal z table, we have,  

P(x > 165) = 1 - 0.8413 = 0.1587 = 15.87\%

Approximately, 159 men weighs more than 165 pounds.

P(men weighing less than 135 pounds)

P(x < 135)

P( x < 135) = P( z < \displaystyle\frac{135 - 150}{15}) = P(z < -1)

Calculation the value from standard normal z table, we have,  

P(x < 135) = 0.1587 = 15.87\%

Approximately, 159 men weighs less than 135 pounds.

6 0
2 years ago
The owner of a toy store received a shipment of 1552 party favors came in 28 boxes the same number of party favors were in 27 of
klemol [59]

Answer : (13)

Let x be the number of party favors in each (27 boxes)

Let y be the number of favors in last box

Given : total of party favors = 1552

27 boxes + last box = 1552

27x + y = 1552

We divide 1552 by 27, the quotient will be our x and the remainder will be y(number of party favors in last box)

Use long division

= 57 and remainder is 13

the number of party favors in each of (27 boxes) = 57

the number of party favors in last box = 13

So, there are 13 party favors in last box.

Read more on Brainly.com - brainly.com/question/9624402#readmore

3 0
2 years ago
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