X2 + x - 6 = 0 A. -2, -3 B. -2, 3 C. 2, -3 D. 2, 3

Malia is planting a garden in her yard.The dimensions are represented by x+4 feet and x+8 feet. She needs fencing to go around t

Kobotan [32]

Answer:

I think you would have to do 2(x+4)+2(x+8) because it it talking about perimeter and if you imagine a square or rectangle, there are four sides, so just adding x+4 and x+8 wont be the whole perimeter. So, if you solve the expression 2(x+4)+2(x+8) it will give you 4x+24 XD

Step-by-step explanation:

4 0

3 years ago

Tom and his bothers caught 100 fish on a weeklong fishing trip. The total weight of the fish was 235 pounds write an expression

gtnhenbr [62]

If the weight of each fish is the same all you need to do is to divide 235 by 100 to get 2.35 pounds per fish.
Hope this helps!

7 0

3 years ago

The diagonal AC is drawn in parallelogram ABCD. Which method

tensa zangetsu [6.8K]

Answer: 1 SSS
Explanation: I hope this helps you

8 0

3 years ago

Read 2 more answers

jose and two of his friends are collecting baseball cards. Their goal is for the three of them to collect at least 100 cards in

Nadya [2.5K]

Answer:

B

Step-by-step explanation:

6 0

3 years ago

Read 2 more answers

A fire company keeps two rescue vehicles. Because of the demand on the vehicles and the chance of mechanical failure, the probab

VashaNatasha [74]

Answer:

(a) P (Both vehicles are available at a given time) = 0.81

(b) P (Neither vehicles are available at a given time) = 0.01

(c) P (At least one vehicle is available at a given time) = 0.99

Step-by-step explanation:

Let A = Vehicle 1 is available when needed and B = Vehicle 2 is available when needed.

<u>Given</u>:

The availability of one vehicle is independent of the availability of the other, i.e. P (A ∩ B) = P (A) × P (B)

P (A) = P (B) = 0.90

(a)

Compute the probability that both vehicles are available at a given time as follows:

P (Both vehicles are available) = P (Vehicle 1 is available) ×

P (Vehicle 2 is available)

$P(A\cap B)=P(A)\times P(B)$

$=0.90\times0.90\\=0.81$

Thus, the probability that both vehicles are available at a given time is 0.81.

(b)

Compute the probability that neither vehicles are available at a given time as follows:

P (Neither vehicles are available) = [1 - P (Vehicle 1 is available)] ×

[1 - P (Vehicle 2 is available)]

$P(A^{c}\cap B^{c})=[1-P(A)]\times [1-P(B)]\\$

$=(1-0.90)\times (1-0.90)\\=0.10\times0.10\\=0.01$

Thus, the probability that neither vehicles are available at a given time is 0.01.

(c)

Compute the probability that at least one vehicle is available at a given time as follows:

P (At least one vehicle is available) = 1 - P (None of the vehicles are available)

$=1-[P(A^{c})\times P(B^{c})]\\=1-0.01.....(from\ part\ (b))\\ =0.99$

Thus, the probability that at least one vehicle is available at a given time is 0.99.

6 0

3 years ago