Mx^{2} + n x^{2} when m = -5 and n = 3 what is the value of the expression

Write and solve a real world word problem to fit the expression -50 ÷ 5.

MissTica

Answer: THe person who rented a apartment had been in debt for 50 dollars, but she didnt have a lot of money so the land lord would divide the rent by 5

Step-by-step explanation:

6 0

2 years ago

If I wanted to estimate the √99, the first step would be to find the two squares that 99 lies on the numberline. I could then th

____ [38]

Answer:

Step-by-step explanation:

Smaller perfect squares near 99 is 81

Larger perfect square near 99 is 100

First step would be to find the two perfect squares that lies between on the number line. I could then think about the number 99 and how close it is to the smaller perfect square and the larger perfect square. That could tell me how far above or below the of the two perfect squares 99 lies on the number line. I could then take the square root of the perfect squares to see how I would estimate the square root of 99. The √99 is almost 10.

81 < 99 < 100

√81 < √99 < √100

8 < √99 < 10

So, √99 is almost 10.

8 0

1 year ago

Read 2 more answers

I need the answer please

lutik1710 [3]

5 inches : 2 feet

÷2.5

120 inches : 48 feet

÷2.5

answer: 48 feet

4 0

3 years ago

The opponents of soccer team A are of two types: either they are a class 1 or a class 2 team. The number of goals team A scores

NikAS [45]

Answer:

a) The expected number of goals team A will score is 5.1

b) The probability that team A will score a total of 5 goals is 0.1147

Step-by-step explanation:

Let X be the amount of goals scored by team A in both matches. Let X1 and X2 be the total amount of goals team A scores in match 1 and 2 respectively, then X = X1+X2, and also

a)

E(X) = E(X1+X2) = E(X1)+E(X2) = 0.6*2+0.4*3 + 0.3*2+0.7*3 = 5.1

b) In order for X to be equal to 5 we have 5 possibilities

- X1 is 0 and X2 is 5

- X1 is 1 and X2 is 4

- X1 is 2 and X2 is 3

- X1 is 3 and X2 is 2

- X1 is 4 and X2 is 1

- X1 is 5 and X2 is 0

Let T1 be a poisson distribution with mean λ = 2, then

$P(T1=0) = e^{-2}$

$P(T1=1) = 2 * e^{-2}$

$P(T1=2) = 2 * e^{-2}$

$P(T1=3) = \frac{4}{3} \, e^{-2}$

$P(T1=4) = \frac{2}{3}\, e^{-2}$

$P(T1=5) = \frac{4}{15}\,e^{-2}$

Lets do the same with a Poisson distribution T2 with mean λ = 3

$P(T2=0) = e^{-3}$

$P(T2=1) = 3 \, e^{-3}\\P(T2=2) = \frac{9}{2} \, e^{-3}\\P(T2=3) = \frac{9}{2} \, e^{-3}\\P(T2=4) = \frac{27}{8} \, e^{-3}\\P(T2=5) = \frac{81}{40} \, e^{-3}$

Now, we are ready to compute the probability that X is equal to 5.

$P(X1 = 0, X2 = 5) = (0.6* e^{-2} + 0.4*e^{-3}) * (0.3*\frac{4}{15}e^{-2} + 0.7*\frac{81}{40} e^{-3}) = 0.00823\\P(X1 = 1, X2 = 4) = (0.6* 2e^{-2} + 0.4*3 e^{-3}) * (0.3*\frac{2}{3}e^{-2} + 0.7*\frac{27}{8} e^{-3}) = 0.03214\\P(X1 = 2, X2 = 3) = (0.6* 2e^{-2} + 0.4*\frac{9}{2}e^{-3}) * (0.3*\frac{4}{3}e^{-2} + 0.7*\frac{9}{2} e^{-3}) = 0.05317\\P(X1 = 3, X2 = 2) = (0.6* \frac{4}{3}e^{-2} + 0.4*\frac{9}{2}*e^{-3}) * (0.3*2e^{-2} + 0.7*\frac{9}{2} e^{-3}) = 0.0471$

$P(X1 = 4, X2 = 1) = (0.6* \frac{2}{3}e^{-2} + 0.4*\frac{27}{8}e^{-3}) * (0.3*2e^{-2} + 0.7*3e^{-3}) = 0.0225\\P(X1 = 5, X2 = 0) = (0.6* \frac{4}{15}e^{-2} + 0.4*\frac{81}{40}e^{-3}) * (0.3*e^{-2} + 0.7*e^{-3}) = 0.0047$

We can conclude that

P(X = 5) = 0.00823+0.03214+0.05317+0.0471+0.0225+0.0047 = 0.1147

The probability that team A will score a total of 5 goals is 0.1147

7 0

3 years ago

Patrick collected 192 seashells at the beach.He made five necklaces with an equal number of shells on each necklace.He used the

Goryan [66]

Answer:

I'm going to put the most shells possible on the necklaces this would be 38, 38x5=190, meaning that we have a remainder of 2.

2 shells would be on the magnet

6 0

2 years ago