A trapezoid has an area of 80 square centimeters. If the trapezoid has a base measuring 8 centimeters and a height of 8 centimeters, what is the length of the other base?

To determine the wall space needed for Mr. Hart's picture, calculate the area of the rectangle formed by the picture's dimensions: length (13 inches) times width (9 inches), resulting in 117 square inches of wall space.

The student's question about Mr. Hart hanging a picture on a wall involves calculating the wall space the picture will occupy. This is essentially a question about finding the area of a rectangle. To find the area, one must multiply the length by the width of the rectangle. In this case, the length of the picture frame is 13 inches and the width is 9 inches. The calculation will be:

Area = Length × Width

Area = 13 inches × 9 inches

Area = 117 square inches

Therefore, the picture will need 117 square inches of wall space.

Answer- C(16,12)

C(n,r) = n! / r!* (n-r)!

So,

C(16,4) = 16!/ 4! * 12!

C(16,12) = !6!/ 12! * 4!

Answer:

2 miles

Step-by-step explanation:

Sara loves to go hiking.

She hiked 3 miles on Monday 4 miles on Tuesdays and 3 miles on Thursday.

Last week Sara hiked 8 miles

This week sara hiked = Monday + Tuesday + Thursday

                                   = 3 + 4 + 3

                                  = 10 miles

     Last week hiked = 8 miles

Difference in two week hiked by Sarah = 10 - 8

                                                                    = 2 miles

So, Sarah hike 2 miles more than last week.

Answer:

B

Step-by-step explanation:

imagine math

What is the interquartile range of this data? 6 7 8 9 Box-and whisker plot titled Temperatures in Celsius ranging from 12 to 48 with the five number summary from left to right 18, 25, 28, 33, and 42

Solution:

The problem is not clear.

But, considering data set is:

18,25,28,33,42

So, if we consider the data set, Interquartile Range is the difference of Quartile3-Quartile1

IQR=Q3-Q1

To find, Q1 and Q3. Let us find the median, or, Q2 first.

Median is the middle term of the data set.

So, Median, Q2=28

Q1 is the median of first half of the data set. That is, 18,25,28

So, Lower Quartile, Q1=25

Q3 is the median of second half of data set. That is, 28, 33,42

So, Upper Quartile, Q3=33

IQR=Q3-Q1

=33-25=8

Answer: IQR=8

If we consider data set: 6,7,8,9

Median, Q2=[tex] \frac{(7+8)}{2}=\frac{15}{2} [/tex]=7.5

Lower Quartile, Q1=6.25

Upper Quartile, Q2=8.75

IQR=Q3-Q1=8.75-6.25=2.5

IQR=2.5

Answer:

Red kite has maximum altitude with 23.1 feet higher.

Step-by-step explanation:

Given : Emma enjoys flying kites. her red kite has a maximum altitude of 117.46 meters, and her black kite has a maximum altitude of 362.26 feet.

one foot is the same as 0.3048 meters.

To find : Which kite has a higher maximum altitude, and how many feet higher is it?

Solution :

First we have to convert meter into feet.

We have given, 1 feet = 0.3048 meters.

[tex]1 \text{ meter }=\frac{1}{0.3048} \text{ feet}[/tex]

So, red kite has a maximum altitude of 117.46 meters

In feet, red kite has a maximum altitude of

[tex]117.46\text{ meter }=\frac{117.46}{0.3048}=385.36 \text{ feet}[/tex]

Now, Red kite has a maximum altitude of 385.36 feet

Black kite has a maximum altitude of 362.26 feet.

385.36>362.26 feet.

Difference is 385.36-362.26=23.1 ft.

Therefore, Red kite has maximum altitude with 23.1 feet higher.

Answer: 11 of each type

Step-by-step explanation:

Using Frequency table, Option: (d) is correct.

Frequency refers to "the number of times an event or values occurs".

According to the question,

The number of deer spotted in preserve is

32,27,15,1,42,35,46,29,38,18,40,38,32,34

The number deer appear in the interval 10 - 19 is 3

The number deer appear in the interval 20 - 29 is 2

The number deer appear in the interval 30 - 39 is 6

The number deer appear in the interval 40 - 49 is 3

From the above data, we can form frequency table

Deer                  10 - 19          20 - 29     30 - 39         40 - 49

Frequency           3                    2              6                    3

Hence, the answer is Option: (d)

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After 4 years, with an initial deposit of $6,000 at a 1.5% annual interest rate compounded monthly, Isabel will have approximately $6,370.07 in her account.

Calculating Compound Interest

To calculate the amount of money Isabel will have in her account after 4 years with an initial deposit of $6,000 at an annual interest rate of 1.5% compounded monthly, we can use the compound interest formula:

V(t) = P(1 + r/n)nt

P = principal amount (initial investment) = $6,000

r = annual interest rate (as a decimal) = 0.015

n = number of times the interest is compounded per year = 12

t = number of years the money is invested or borrowed for = 4

Substituting these values into the formula, we get:

V(4) = $6,000(1 + 0.015/12)12(4)

This simplifies to:

V(4) = $6,000(1 + 0.00125)48

Calculating the value inside the parentheses first:

1 + 0.00125 = 1.00125

Then raising it to the 48th power gives us:

1.0012548 ≈ 1.061678

Multiplying this by the principal:

V(4) ≈ $6,000 × 1.061678 ≈ $6,370.07

After 4 years, Isabel's account will have approximately $6,370.07.

Answer:$6,370.07

Step-by-step explanation:

The probability that the result is a multiple of 5 and a multiple of 2 is 6/11

Suppose that there are finite elementary events in the sample space of the considered experiment, and all are equally likely.

Then, suppose we want to find the probability of an event E.

Then, its probability is given as

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{n(E)}{n(S)}[/tex]

where favorable cases are those elementary events who belong to E, and total cases are the size of the sample space.

For this case, we're given that:

Take E = Event of spinner's spin resulting in multiple of 5 or multiple of 2

S = results of Sample

Then, favorable results (in favor of E) are: 2,4,5,6,8,10 (total 6 in count)

All possible results are 1, 2, ... , 11  (total 11 in count)

Thus, we get:

[tex]P(E) = \dfrac{\text{Number of favorable cases}}{\text{Number of total cases}} = \dfrac{6}{11}[/tex]

Thus, the probability that the result is a multiple of 5 and a multiple of 2 is 6/11

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In a spinner numbered 1 through 11, there is only one number (10) that is both a multiple of 5 and a multiple of 2. As there are 11 possible outcomes in a single spin, the probability of spinning and landing on 10 is 1/11.

The subject of this question is mathematical probability. Given that there is a spinner with 11 equal areas numbered 1 through 11, we are asked to find the probability of landing on a number that is both a multiple of 5 and a multiple of 2 when the spinner is spun once.

Firstly, we identify the numbers which are both multiples of 5 and 2 between 1 and 11. In this case, only one such number exists, which is 10. Hence, there is only one favorable outcome. The total number of outcomes are 11 (as the spinner has 11 equal areas).

Probability is defined as 'number of favorable outcomes' divided by the 'total number of outcomes'. In this case, since there is 1 favorable outcome (i.e. landing on '10') and 11 possible outcomes (any of the numbers 1 to 11), the probability is therefore 1/11.

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The solution is: The box plot is attached.

In statistics and probability theory, the median is the value separating the higher half from the lower half of a data sample, a population, or a probability distribution. For a data set, it may be thought of as "the middle" value.

We first order the data from least to greatest:

112, 116, 134, 134, 135, 141, 149, 154, 156, 156

We find the median (middle value).  There are 10 data values, which puts the median between 135 and 141:

(135+141)/2 = 276/2 = 138

The Upper Quartile (UQ) is the median of the upper half of data, cut by the median.  This is 154.

The Lower Quartile (LQ) is the median of the lower half of data, cut by the median.  This is 134.

The middle line of the box is drawn at the median, 138.  The left side of the box is at the LQ, 134.  The right side of the box is at the UQ, 154.  There is a whisker drawn from the right side to the highest value, 156.  There is a whisker drawn from the left side to the smallest value, 112.

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The probability that when 12 adults are randomly selected, 3 or fewer are in excellent health is 0.947.

To find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health, we can use the binomial distribution formula.

The formula for the probability of getting exactly k successes in n trials, with a probability of success p, is:

P(X = k) = (n choose k) * (p^k) * ((1-p)^(n-k))

In this case, n = 12, k = 3, and p = 0.38 (as 38% of adults reported excellent health).

So, the probability that 3 or fewer adults are in excellent health can be calculated as follows:

P(X <= 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)

Let's calculate each term:

Adding up these probabilities:

P(X <= 3) = 0.190 + 0.342 + 0.269 + 0.146 = 0.947

Rounded to three decimal places, the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health is 0.947.

The probable question may be:

In a study, 38% of adults questioned reported that their health was excellent. A researcher wishes to study the health of people living close to a nuclear power plant. Among 12 adults randomly selected from this area, only 3 reported that their health was excellent. Find the probability that when 12 adults are randomly selected, 3 or fewer are in excellent health. Round to three decimal places.