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vampirchik [111]
4 years ago
14

Kristin weighs three kittens at a vet's clinic. The heaviest one weighs 3.28 pounds. The heaviest kitten is 1.056 pounds heavier

than the medium-weight one. The lightest kitten is 1.2 pounds lighter than the medium-weight kitten. What is the total weight of the kittens?
Mathematics
2 answers:
ehidna [41]4 years ago
6 0

Answer:

Total weight of all kittens = 6.528 pounds

Step-by-step explanation:

If there are 3 kittens, kitten (1), kitten (2), kitten(3).

kitten (1) is the heaviest.

kitten (2) is the medium weight.

kitten (3) is the lightest.

Weight kitten (1) = 3.28 pounds

Weight of kitten (1) - weight of kitten (2) = 1.056 pounds

3.28 - weight of kitten (2) = 1.056

Weight of kitten (2) = 3.28 - 1.056 = 2.224 pounds

Weight of kitten (2) - weight of kitten (3) = 1.2 pounds

2.224 - weight of kitten (3) = 1.2

weight of kitten (3) = 2.224 - 1.2 = 1.024 pounds

Total weight of all kittens = 3.28 + 2.224 + 1.024 = 6.528 pounds.

iogann1982 [59]4 years ago
4 0

Answer:

6.704

Step-by-step explanation:

you will talk 1.056 and subtract it from 3.28 to get the weight of the medium one. which would be 2.224. then you would add them all up. 3.28 + 2.224 + 1.2 and it will get you 6.704

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A card is drawn from a deck of 10 cards numbered 1 through 10 and a number cube is rolled. Find each probability.
Helen [10]

Answer:

1. P (10 and 3) = 1/10 * 1*6 = 1/60.

2. P (two even numbers)  = 5/10 * 3/6 = 15/60 = 1/4.

3. P (two prime numbers)  = 4/10 * 3/6 = 12/60 = 1/5.

4. P (9 and an odd number)  = 1/10 * 3/6 = 3/60 = 1/20.

5. P (two numbers less than 4)  = 3/10 * 3/6 = 9/60 = 3/20.

6. P (two numbers greater than 5) = 5/10 * 1/6 = 5/60 = 1/12.

Step-by-step explanation:

The card has the numbers 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10.

The number cube has the numbers 1, 2, 3, 4, 5, and 6

In the deck of cards, 5 are odd and 5 are even. Also, there are 4 prime numbers in the deck. Also, there are 3 even and 3 odd numbers on the number cube. Also, there are 3 prime numbers cube. Therefore,

P(Selecting any card from the deck) = 1/10.

P(Any number is rolled on the dice) = 1/6.

Assume that probability of the cards and the cube are independent. So the probabilities can be calculated.

1. P (10 and 3) = 1/10 * 1*6 = 1/60.

2. P (two even numbers)  = 5/10 * 3/6 = 15/60 = 1/4.

3. P (two prime numbers)  = 4/10 * 3/6 = 12/60 = 1/5.

4. P (9 and an odd number)  = 1/10 * 3/6 = 3/60 = 1/20.

5. P (two numbers less than 4)  = 3/10 * 3/6 = 9/60 = 3/20.

6. P (two numbers greater than 5) = 5/10 * 1/6 = 5/60 = 1/12.

So the probabilities have been mentioned!!!

3 0
3 years ago
Find the area of the shaded region.
Anvisha [2.4K]

Answer:

73.4 cm^2

Step-by-step explanation:

The equation is :

πr^2

Substitute the radius in:

π(5)^2

Solve:

Area of full circle = 25π

(Note : leave this answer in terms of 'pi' so it is easier to handle

Next, find the area of the sector

The equation for this is:

(angle/360) x πr^2

Substitute the values in:

(80/360) x π(5)^2

Solve :

Area of sector = (50/9)π

Now, find the area of the triangle:

1/2 absinC

Substitute the values in:

1/2(5)(5) x sin(80) = 12.31009691

Subtract this answer from the area of the sector

Answer = 5.14319607

Subtract this from the area of the whole circle

Answer = 73.39662073

To the nearest tenth, that would be 73.4 cm^2

Mark brainliest pls

3 0
3 years ago
How many squares to shade
slamgirl [31]
You got to shade five squares
6 0
4 years ago
Question 6 of 12<br> ✓ 49<br> A. 8<br> O O<br> B. 14<br> C. 7<br> SUBMIT
amm1812

Answer:

C. 7

Step-by-step explanation:

7 · 7 = 49

I hope this helps!

3 0
3 years ago
A baby's crib has an area of 690:square inches and a perimeter of 106 inches.What are the dimensions of the crib
KengaRu [80]

Answer:

The crib is 30 inches wide and 23 inches long.

Step-by-step explanation:

Assuming that the crib is rectangular, then the crib has a length L and a width W.

And remember that for a rectangle of width W and length L, the perimeter is:

P = 2*L + 2*W = 2*(L + W)

And the area is:

A = L*W

In this case, the perimeter is 106 in, then:

P = 106in = 2*(L + W)

And we also know that the area is 690 in^2

Then:

A = 690in^2 = L*W

Then we have a system of equations:

106in = 2*(L + W)

690in^2 = L*W

To solve this, first, we need to isolate one of the variables in one of the equations.

Let's isolate L in the first one:

106in/2 = L + W

53 in = L + W

53in - W = L

Now we can replace this in the other equation:

690in^2 = L*W = (53in - W)*W = 53in*W - W^2

690in^2 =  53in*W - W^2

Now we need to solve this for W.

This is a quadratic equation:

W^2 - 53in*W + 690 in^2 = 0

The solutions of this equation are given by Bhaskara's formula:

W = \frac{-(-53in)\pm \sqrt{(-53in)^2 - 4*1*(690in^2)}  }{2} =  \frac{57in \pm 7in}{2}

Then the two possible solutions of W are:

W = (53in + 7in)/2 = 30in

W = (53in - 7in)/2 = 23in

We can choose any one of these, so let's choose W = 30in

If we replace this in the equation: "53in - W = L"

We can find the value of L:

53in - 30in = 23in = L

Then we have:

W = 30in

L = 23in

4 0
3 years ago
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