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3 years ago

WHOEVER ANSWERS BOTH QUESTIONS FIRST WILL GET MARKED THE BRAIN THINGY (I'll also check to see if you were right or not)

Mathematics

2 answers:

svet-max [94.6K]3 years ago

8 0

Help me wit this one too pls If an image is translated 2 units to the right and 5 units down which of these algebraic representation is the right one? Explain why you selected that answer A. (x-2, y+5) B. (x+5, y-2) C. (x-5, y+2) D. (x+2, y-5)

n200080 [17]3 years ago

3 0

V≈28730.91 and if the radius was 18 it would be V≈24429.02. also ur second answer is correct. btw there are calculators online for finding volume in spheres wayyy easier & fast

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Name a 2 digit odd number that is prime.name a 2 digit odd number that is compostite

e-lub [12.9K]

Prime for 2 digit odd number: 13 , 11 , 17, etc

Odd number that is composite: 39, 25, 15, etc

Hope this helps

7 0

3 years ago

Janae wrote the following polynomial expression: 2x5 - 4x3 + 6x8.

emmasim [6.3K]

Answer:

Justin is correct

Step-by-step explanation:

2x^5 - 4x^3 + 6x^8.

We need to rewrite the polynomial in decreasing order

6x^8 +2x^5 -4x^3

The leading coefficient is 6

8 0

3 years ago

A box contains 3 red pencils, 4 blue pencils, 2 green pencils, and 1 regular pencil.

raketka [301]

Answer:

100

Step-by-step explanation:

add all the number together 3 + 4 + 2 +1=10

250/10=25

then times 25 with the 4 number at the top

25 times 4 =100

so there is 100 blue pencils

8 0

3 years ago

One household is to be selected at random from a town. The probability that the household has a cat is 0.20.2 . The proba

dimulka [17.4K]

Answer:

There is a 50% probability that the household has a dog, given that the household has a cat.

Step-by-step explanation:

We solve this problem building the Venn's diagram of these probabilities.

I am going to say that:

A is the probability that a household has a cat.

B is the probability that a household has a dog.

We have that:

$A = a + (A \cap B)$

In which a is the probability that a household has a cat but not a dog and $A \cap B$ is the probability that a household has both a cat and a dog.

By the same logic, we have that:

$B = b + (A \cap B)$

The probability that the household has a cat or a dog is 0.5

$a + b + (A \cap B) = 0.5$

The probability that the household has a dog is 0.4

$B = 0.4$

$B = b + (A \cap B)$

$b = 0.4 - (A \cap B)$

The probability that the household has a cat is 0.2.

$A = 0.2$

$A = a + (A \cap B)$

$a = 0.2 - (A \cap B)$

$a + b + (A \cap B) = 0.5$

$0.2 - (A \cap B) + 0.4 - (A \cap B) + (A \cap B) = 0.5$

$A \cap B = 0.1$

What is the probability that the household has a dog, given that the household has a cat?

20% of the households have a cat, and 10% have both a cat and a dog. So

$P = \frac{A \cap B}{A} = {0.1}{0.2} = 0.5$

There is a 50% probability that the household has a dog, given that the household has a cat.

4 0

3 years ago

Please help me ‍♀️ im a tad bit slow-

lions [1.4K]

9514 1404 393

Answer:

Michelle
$5

Step-by-step explanation:

The dollar value of the 20% coupon will increase if the discount is from a larger amount. Applying the $25 coupon first reduces the amount that the 20% applies to, so reduces the dollar value of the 20% coupon.

Michelle used the 20% coupon first, so its value was 20%($100) = $20.

Natasha used the $25 coupon first, so the value of the 20% coupon was ...

20%($100 -25) = 20%(%75) = $15.

Michelle paid $20 -15 = $5 less.

5 0

3 years ago