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3 years ago

What is the area of a trapezoid with bases of 15.8 yd and 21.8 yd and a height of 11.7 yd?

Mathematics

2 answers:

DedPeter [7]3 years ago

7 0

Answer:

B) 219.96 yd2

Step-by-step explanation:

The formula for finding the area of a trapezoid is:

(a + b) /2*h

Now you would plug in the numbers

(15.8 + 21.8) /2*11.7

Now solve this in order of operations

(15.8 + 21.8) /2*11.7

37.6/2*11.7

18.8*11.7

219.96

DedPeter [7]3 years ago

3 0

Add the bases together, divide that by 2 then multiply by the height.

15.8 + 21.8 = 37.6

37.6/2 = 18.8

18.8 x 11.7 = 219.96 yd^2

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Fundraising The robotics team needs new uniforms. The students plan to sell plush toy lions (the school mascot) for $5 each.

nataly862011 [7]

Answer:

Company C!

Step-by-step explanation:

In order to see which company has the best buy we have to see the amount each company sells for one stuffed mascot. In order to do this we have to divide .

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Company B sells 18 lions for $50.22 this is the same as 1 stuffed lion for $2.79.

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Thus as you can see the company with the best buy is company C.

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4 years ago

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Find an equation in rectangular coordinates for the cylindrical equation r = 2sintheta

frutty [35]

Answer:

$x^2+(y-1)^2=1$

Step-by-step explanation:

The given cylindrical equation is

$r=2\sin \theta$

Multiply both sides by r.

$r^2=2r\sin \theta$ .... (1)

The required formulas are

$x=r\cos \theta,y=r\sin \theta,r^2=x^2+y^2$

Substitute $r\sin \theta =y,r^2=x^2+y^2$ in equation (1).

$x^2+y^2=2y$

$x^2+y^2-2y=0$

Add 1 on both sides.

$x^2+(y^2-2y+1)=1$

$(x-0)^2+(y-1)^2=1^2$ $[\because (a-b)^2=a^2-2ab+b^2]$

It is the equation of a circle centered at (0,1) with radius 1.

Therefore, the equation in rectangular coordinates is $x^2+(y-1)^2=1$ .

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4 years ago

At a 3-mile cross-country race, an athlete runs 2 miles at 6 mph and 1 mile at 12 mph. What is the athlete's average speed?

Tom [10]

Answer: the average speed is 7.5 mph or at least im pretty sure

Step-by-step explanation:

This is because if you make the 2 miles at 6 mph to 1 mile at 3 mph then you get 1 mile at 3mph and 1 mile at 12 mph. now that you have the unit rates of these you can add them. 1 + 1 = 2 miles and 3 + 12 = 15 mph now divide this by two and you get 7.5 mph per mile.

I hope this helped, if it did could you give me a good rating? lol

8 0

3 years ago

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. Consider the sample m

love history [14]

Answer:

a) The expected value of the sample mean weight is 20.4 pounds.

b)The standard deviation of the sample mean weight is 0.123.

c) There is a 14.46% probability the sample mean weight will be less than 20.27.

d) This value is $c = 20.6153$ .

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$ .

Problems of normally distributed samples can be 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.

In this problem, we have that:

A particular variety of watermelon weighs on average 20.4 pounds with a standard deviation of 1.23 pounds. This means that $\mu = 20.4, \sigma = 1.23$

Consider the sample mean weight of 100 watermelons of this variety. This means that $n = 100$ .

a. What is the expected value of the sample mean weight? Give an exact answer.

By the Central Limit Theorem, it is the same as the mean of the population. So the expected value of the sample mean weight is 20.4 pounds.

b. What is the standard deviation of the sample mean weight? Give your answer to four decimal places.

By the Central Limit Theorem, that is:

$s = \frac{\sigma}{\sqrt{n}} = \frac{1.23}{\sqrt{100}} = 0.123$

The standard deviation of the sample mean weight is 0.123.

c. What is the approximate probability the sample mean weight will be less than 20.27?

This is the pvalue of Z when $X = 20.27$ .

Since we are working with the sample mean, we use $s$ instead of $\sigma$ in the Z score formula

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

$Z = \frac{20.27 - 20.4}{0.123}$

$Z = -1.06$

$Z = -1.06$ has a pvalue of 0.1446.

This means that there is a 14.46% probability the sample mean weight will be less than 20.27.

d. What is the value c such that the approximate probability the sample mean will be less than c is 0.96?

This is the value of $X = c$ that is in the 96th percentile, that is, it's Z score has a pvalue 0.96.

So we use $Z = 1.75$

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

$1.75 = \frac{c - 20.4}{0.123}$

$c - 20.4 = 0.123*1.75$

$c = 20.6153$

This value is $c = 20.6153$ .

6 0

3 years ago

Pls get it right it’s a test

Nady [450]

Answer:

77.98

is not approximately

is?

Step-by-step explanation:

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3 years ago

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