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2 years ago

1 All sides are congruent. Paralelogram Rhombus Rectangle Square

Mathematics

1 answer:

Amanda [17]2 years ago

8 0

Answer:

Rhombus

Square

Step-by-step explanation:

Rhombuses and squares have sides that are always congruent.

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Mrs. Lang is 4 times as old as her daughter Jill. The sum of their ages is 60 years. Set up the two equations that can be used t

KatRina [158]

Answer:

The two equations that can be used to find each of their ages are $x=4 \times y$ and $x+y=60$ .

Step-by-step explanation:

We are given that Mrs. Lang is 4 times as old as her daughter Jill. The sum of their ages is 60 years.

Let the age of Mrs. Lang be 'x years' and the age of her daughter Jill be 'y years'.

Now, according to the question;

The first condition states that Mrs. Lang is 4 times as old as her daughter Jill, that means;

$x=4 \times y$ ----------------- [equation 1]

The second condition states that the sum of their ages is 60 years, that means;

$x+y=60$

$4y + y = 60$ {using equation 1}

$5y=60$

$y=\frac{60}{5}$

y = 12 years

Now, putting the value of y in equation 1 we get;

$x=4 \times y$

$x= 4 \times 12$ = 48 years

Hence, the age of Mrs. Lang is 48 years and her daughter Jill is 12 years old.

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3 years ago

Write the fraction 4/10 as a product of a whole number and a unit fraction

iren2701 [21]

Answer:

Step-by-step explanation:

6 0

3 years ago

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Find the perimeter of the rectangle whose coordinates are A (5, 3), B (8, 3), C

MaRussiya [10]

Answer:

Step-by-step explanation:

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3 years ago

A team of 10 players is to be selected from a class of 6 girls and 7 boys. Match each scenario to its probability. You have to d

tankabanditka [31]

The selection of r objects out of n is done in

$C(n, r)= \frac{n!}{r!(n-r)!}$ many ways.

The total number of selections 10 that we can make from 6+7=13 students is

$C(13,10)= \frac{13!}{3!(10)!}= \frac{13*12*11*10!}{3*2*1*10!}= \frac{13*12*11}{3*2}= 286$
thus, the sample space of the experiment is 286

A.
"The probability that a randomly chosen team includes all 6 girls in the class."

total number of group of 10 which include all girls is C(7, 4), because the girls are fixed, and the remaining 4 is to be completed from the 7 boys, which can be done in C(7, 4) many ways.

 $C(7, 4)= \frac{7!}{4!3!}= \frac{7*6*5*4!}{4!*3*2*1}= \frac{7*6*5}{3*2}=35$

P(all 6 girls chosen)=35/286=0.12

B.
"The probability that a randomly chosen team has 3 girls and 7 boys."

with the same logic as in A, the number of groups were all 7 boys are in, is

 $C(6, 3)= \frac{6!}{3!3!}= \frac{6*5*4*3!}{3!3!}= \frac{6*5*4}{3*2*1}=20$

so the probability is 20/286=0.07

C.
"The probability that a randomly chosen team has either 4 or 6 boys."

case 1: the team has 4 boys and 6 girls

this was already calculated in part A, it is 0.12.

case 2, the team has 6 boys and 4 girls.

there C(7, 6)*C(6, 4) ,many ways of doing this, because any selection of the boys which can be done in C(7, 6) ways, can be combined with any selection of the girls.

 $C(7, 6)*C(6, 4)= \frac{7!}{6!1}* \frac{6!}{4!2!} =7*15= 105$

the probability is 105/286=0.367

since case 1 and case 2 are disjoint, that is either one or the other happen, then we add the probabilities:

0.12+0.367=0.487 (approximately = 0.49)

D.
"The probability that a randomly chosen team has 5 girls and 5 boys."

selecting 5 boys and 5 girls can be done in

 $C(7, 5)*C(6,5)= \frac{7!}{5!2} * \frac{6!}{5!1}=21*6=126$

many ways,

so the probability is 126/286=0.44

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4 years ago

Read 2 more answers

ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately no

dimaraw [331]

Answer: 5.37

Step-by-step explanation:

Let x = ACT scores.

Given: ACT scores have a mean of 20.8 and 9 percent of the scores are above 28. The scores have a distribution that is approximately normal.

i.e. P(X>28)=0.09 (i)

Now,

$P(X>28)=P(\dfrac{X-\mu}{\sigma}>\dfrac{28-20.8}{\sigma})\\\\= P(z>\dfrac{7.2}{\sigma})\ \ \ \ [z=\dfrac{X-\mu}{\sigma}]$ (ii)

One -tailed z value for p-value of 0.09 =1.3408 [By z-table]

From (i) and (ii)

$\dfrac{7.2}{\sigma}=1.3408\\\\\Rightarrow\ \sigma=\dfrac{7.2}{1.3408}\\\\\Rightarrow\ \sigma=5.37$

Hence, the standard deviation = 5.37

5 0

3 years ago

1 All sides are congruent. Paralelogram Rhombus Rectangle Square​

1 All sides are congruent. Paralelogram Rhombus Rectangle Square