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3 years ago

I need help can anyone help me

Mathematics

1 answer:

kolbaska11 [484]3 years ago

6 0

So for each day, Tessa ran 2.25 miles. So, multiply 2.25 miles by the number of days, which is 14, to get the total number of miles ran.

2.25 x 14 = 31.5 miles

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If you run 6. Miles an hour how long will it take to do 18 miles help quick pls

lys-0071 [83]

18miles/6mph = 3hrs
3 hours

6 0

3 years ago

mr and mrs gracia took their three children to see a matinee on saturday. they spent a total of 55.50, which included 29.95 at t

andrezito [222]

First, you would subtract 29.95 from 55.50, and you get 25.55. because 5 people went, and the cost for each was the same, you would divide 25.55 by five, and get $5.11 per person.

3 0

3 years ago

Simplify polynomial

Kay [80]

Answer:

j² - 5j²k - 2

Step-by-step explanation:

3j² - j²k - 6 - 4j²k - 2j² + 4

To simplify this polynomial, we can collect like terms. A term is number(s) or variable(s) that are grouped together by multiplication. Like terms have the same variable and exponent.

We have three groups of like terms:

The j-squares (j²), the j-squared k (j²k) and the constants (no variable).

Remember to include the negatives!

The j-squares are: 3j² ; -2j²

The j-squares k are: - j²k ; - 4j²k

The constants are: - 6 ; 4

Simplify:

3j² - j²k - 6 - 4j²k - 2j² + 4

Rearrange the polynomial by like terms

= (- j²k - 4j²k) + (3j² - 2j²) + (- 6 + 4)

Add or subtract the like terms

= (-5j²k) + (j²) + (-2)

Remove brackets and rearrange so the negative is not first

= j² + - 5j²k + - 2

Simplify where two signs are together. Adding a negative is subtraction.

= j² - 5j²k - 2 Simplified

8 0

3 years ago

How do I solve this? I need help

Sedaia [141]

Answer: √51
—————————
a^2 + b^2 = c^2
a^2 + 7^2 = 10^2
a^2 + 49 = 100
a^2 = 51
√(a^2) = √51
a = √51

7 0

2 years ago

For about $1 billion in new space shuttle expenditures, NASA has proposed to install new heat pumps, power heads, heat exchanger

11111nata11111 [884]

Answer:

The probability of one or more catastrophes in:

(1) Two mission is 0.0166.

(2) Five mission is 0.0410.

(3) Ten mission is 0.0803.

(4) Fifty mission is 0.3419.

Step-by-step explanation:

Let X = number of catastrophes in the missions.

The probability of a catastrophe in a mission is, P (X) = $p=\frac{1}{120}$ .

The random variable X follows a Binomial distribution with parameters n and p.

The probability mass function of X is:

$P(X=x)={n\choose x}\frac{1}{120}^{x}(1-\frac{1}{120})^{n-x};\x=0,1,2,3...$

In this case we need to compute the probability of 1 or more than 1 catastrophes in n missions.

Then the value of P (X ≥ 1) is:

P (X ≥ 1) = 1 - P (X < 1)

= 1 - P (X = 0)

$=1-{n\choose 0}\frac{1}{120}^{0}(1-\frac{1}{120})^{n-0}\\=1-(1\times1\times(1-\frac{1}{120})^{n-0})\\=1-(1-\frac{1}{120})^{n-0}$

(1)

Compute the compute the probability of 1 or more than 1 catastrophes in 2 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{2-0}=1-0.9834=0.0166$

(2)

Compute the compute the probability of 1 or more than 1 catastrophes in 5 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{5-0}=1-0.9590=0.0410$

(3)

Compute the compute the probability of 1 or more than 1 catastrophes in 10 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{10-0}=1-0.9197=0.0803$

(4)

Compute the compute the probability of 1 or more than 1 catastrophes in 50 missions as follows:

$P(X\geq 1)=1-(1-\frac{1}{120})^{50-0}=1-0.6581=0.3419$

6 0

3 years ago