If the rectangle perimeter equals 42 meters what is the length?

HELP ME PLEASE!!!!<br> ΔAXY ~ ΔABC. What is the value of BC

emmasim [6.3K]

Answer:

BC = 28 $\frac{1}{3}$ ft

Step-by-step explanation:

Given that the triangles are similar then the ratios of corresponding sides are equal, that is

$\frac{AX}{AB}$ = $\frac{XY}{BC}$ , substitute values

$\frac{12}{20}$ = $\frac{17}{BC}$ ( cross- multiply )

12BC = 340 ( divide both sides by 12 )

BC = $\frac{340}{12}$ = $\frac{85}{3}$ = 28 $\frac{1}{3}$

6 0

3 years ago

Q: Tom would like to take out a secured loan to help pay for a vacation this summer. He has offered his car as collateral. His c

givi [52]

Answer:

hi

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3 0

2 years ago

Dante sold 413 cups of lemonade last weekend. George sold 297 cups of lemonade last weekend. How many more cups of lemonade did

TEA [102]

Answer:dante sold 116 more cups.

Step-by-step explanation: subtract 297 from 413

6 0

3 years ago

In a MBS first year class, there are three sections each including 20 students. In the first section, there are 10 boys and 10 g

KIM [24]

Answer:

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

Step-by-step explanation:

The selection is from a sample without replacement, so we use the hypergeometric distribution to solve this question.

Hypergeometric distribution:

The probability of x sucesses is given by the following formula:

$P(X = x) = h(x,N,n,k) = \frac{C_{k,x}*C_{N-k,n-x}}{C_{N,n}}$

In which:

x is the number of sucesses.

N is the size of the population.

n is the size of the sample.

k is the total number of desired outcomes.

Combinations formula:

$C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

All girls from the first group:

20 students, so $N = 20$

10 girls, so $k = 10$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_1 = P(X = 5) = h(5,20,5,10) = \frac{C_{10,5}*C_{10,5}}{C_{20,5}} = 0.0163$

All girls from the second group:

20 students, so $N = 20$

5 girls, so $k = 5$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_2 = P(X = 5) = h(5,20,5,5) = \frac{C_{5,5}*C_{15,5}}{C_{20,5}} = 0.00006$

All girls from the third group:

20 students, so $N = 20$

8 girls, so $k = 8$

5 students will be selected, so $n = 5$

We want all of them to be girls, so we find P(X = 5).

$P_3 = P(X = 5) = h(5,20,5,8) = \frac{C_{8,5}*C_{12,5}}{C_{20,5}} = 0.0036$

All 15 students are girls:

Groups are independent, so we multiply the probabilities:

$P = P_1*P_2*P_3 = 0.0163*0.00006*0.0036 = 3.52 \times 10^{-9}$

$3.52 \times 10^{-9} = 3.52 \times 10^{-7}\%$ probability that all the 15 students selected are girls

7 0

3 years ago

Analyze the data represented in the graph and select the appropriate model. A) exponential B) linear C) none D) quadratic

Karo-lina-s [1.5K]

Answer:

exponentialIt couldn’t be liner, liner is straight, the line is wriggly,, quadriceps obv no, so it has to be exponential.

7 0

3 years ago