All categories

4 years ago

Find the area of the figure. PLZ HURRY

Mathematics

2 answers:

Alex4 years ago

4 0

Answer:

132 cm²

Step-by-step explanation:

Area = rectangle + trapezoid

(7 × 4) + [½(7 + 19) × 8]

28 + 204

132 cm²

Serjik [45]4 years ago

3 0

Answer:

132 cm²

Step-by-step explanation:

This is a composite figure, so we want to find the area of the top rectangle and the bottom trapezoid and then add those together.

The area of a rectangle is denoted by: A = lw, where l is the length and w is the width. Here, the length is l = 4 and the width is w = 7, so the area is:

A = lw

A = 7 * 4 = 28 cm²

The area of a trapezoid is denoted by: A = $\frac{(b_1+b_2)*h}{2}$ , where b1 and b2 are the bases and h is the height. Here, the two bases are 7 and 19 and the height is h = 8. So, the area is:

A = $\frac{(b_1+b_2)*h}{2}$

A = $\frac{(7+19)*8}{2}=26 *4=104$ cm²

Add these together:

28 + 104 = 132 cm²

You might be interested in

Two marbles are drawn without replacement from a box with 3 white, 2 green, 2 red, and 1 blue marble. find the probability that

Kaylis [27]

<span>3/28 exactly, or approximately 10.7% Since we're drawing without replacement, let's first calculate the probability that the 1st marble is white. There are 3 white marbles out of a total of 8 marbles. So the probability of the 1st marble being white is 3/8. Now assuming that the first marble picked is white, the probability of the second marble is 2 out of the 7 remaining marbles. So it's probability is 2/7. This means that the probability of both marbles being white is 3/8 * 2/7 = 6/56 = 3/28, or approximately 10.7%</span>

5 0

3 years ago

The drama club was selling ticketsto the school play. Adult ticketscost $8.00 each, and studenttickets cost $5.00 each. The litt

BlackZzzverrR [31]

We are given a problem that can be solved using a system of linear equations. Let A, be the number of adults, and S the number of students. Since there are in total 142 people and there were two days, this means that the sum of the number of adults and the number of students must be 284, which can be written mathematically as follows:

$A+S=284,(1)$

This is our first equation. The second equation is found using the total sales of $1948. Since the ticket per adult is $8 and per student is $5, we have the following equations:

$8A+5S=1948,(2)$

To solve this equation we will solve for A in equation (1), by subtracting S to both sides;

$\begin{gathered} A+S-S=284-S \\ A=284-S \end{gathered}$

Now we will replace this value in equation (2):

$8(284-S)+5S=1948$

Now we will apply the distributive property:

$2272-8S+5S=1948$

Addins like terms:

$2272-3S=1948$

Subtracting 2272 to both sides;

$\begin{gathered} 2272-2272-3S=1948-2272 \\ -3S=-324 \end{gathered}$

Dividing both sides by -3:

$S=-\frac{324}{-3}=108$

Now we replace this value in equation (1), where we have already solved for A:

$\begin{gathered} A=248-108 \\ A=140 \end{gathered}$

Therefore, there were sold 108 student tickets and 140 adult tickets.

3 0

1 year ago

456÷368<br> what is the answer

Paul [167]

<h3>Answer :</h3>

$\tt 456 \div 368$

$\tt = \dfrac{456}{368}$

$\tt = \dfrac{\cancel{456}^{228}}{\cancel{368}_{184}}$

$\tt = \dfrac{\cancel{228}^{114}}{\cancel{184}_{92}}$

$\tt = \dfrac{\cancel{114}^{57}}{\cancel{92}_{46}}$

$\tt = \dfrac{57}{46}$

$\pink{\tt \therefore \: Answer \: is \: \dfrac{57}{46} \: or \: 1.24 (approx.)}$

7 0

3 years ago

Read 2 more answers

A necklace calls for red, blue and white beads in the ratio of 5: 8:12. Ming has used

notka56 [123]

Answer:

Step-by-step explanation:

find unity

350 : (5 + 8 + 12) =

350 : 25 = 14

multiply the units by the white ratio

14 * 12 = 168

6 0

2 years ago

For the accompanying data set, draw a scatter diagram of the data.x 2 6 6 7 9y 3 2 6 9 5B. by hand compute the correlation coeff

zhenek [66]

Answer:

Step-by-step explanation:

Hello!

Given the data corresponding the variables:

x 2 6 6 7 9

y 3 2 6 9 5

The Scatterplot is attached.

To compute the correlation coefficient you need several auxiliary calculations:

∑X= 30

∑X²= 206

∑Y= 25

∑Y²= 155

∑XY= 162

$r= \frac{sumXY-\frac{(sumX)(sumY)}{n} }{\sqrt{[sumX^2-\frac{(sumX)^2}{n} ][sumY^2-\frac{(sumY)^2}{n} ]} }$

$r= \frac{162-\frac{(30)*(25)}{5} }{\sqrt{[206-\frac{(30)^2}{5} ][155-\frac{(25)^2}{5} ]} }$

r= 0.429 ≅ 0.43

The critical value for r has n-2 degrees of freedom, let's say for example you have α:0.05

$r_{n-2;\alpha } = r_{3;0.05 } = 0.878$

For a two-tailed test.

Because the correlation coefficient is <u>positive</u> and the absolute value of the correlation coefficient, _<u>0.43</u>___, is <u>not greater</u> than the critical value for this data set,_<u>0.878</u>__, <u>no </u>linear relationship exists between x and y.

I hope it helps!

4 0

3 years ago