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2 years ago

Please Please Help Ne In This Math homework?!?!?!

Mathematics

1 answer:

weeeeeb [17]2 years ago

8 0

6 is yes because they have the same angles and lengths while 7 is no because the angles are not the same

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Alonso went to the market with \$55$55dollar sign, 55 to buy eggs and sugar. He knows he needs a package of 121212 eggs that cos

eduard

Answer:

2.75+11.50S $\leq$ 55

Step-by-step explanation:

7 0

2 years ago

A product is composed of four parts. In order for the product to function properly in a given situation, each of the parts must

svet-max [94.6K]

Answer:

P(working product) = .99*.99*.96*.96 = .0.903

Step-by-step explanation:

For the product to work, all four probabilities must come to pass, so that

P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

where

P(Part-1) = 0.96

P(Part-2) = 0.96

P(Part-3) = 0.99

P(Part-4) = 0.99

As all parts are independent, so the formula is P(A∩B) = P(A)*P(B)

P (Working Product) = P(Part-1)*P(Part-2)*P(Part-3)*P(Part-4)

P (Working Product) = 0.96*0.96*0.96*0.99*0.99

P(Working Product) = 0.903

3 0

2 years ago

Estimate the product 8.1 x 4.2

Korvikt [17]

Round 8.1 down to 8
Round 4.2 down to 4

8(4)= 32

The estimated product is 32.
Hope this helps!

6 0

2 years ago

What is the answer please

Marina86 [1]

Hey there!

Use The Pythagorean theorem. It’s a^2+b^2=c^2. Plug in 11 and 60. The equation would be 11^2+60^2=c^2. Simplify to get 121+3600=c^2. Simplify again to get 3721=c^2. Sqaure root each side to get c=61. The answer is the last one, c=61.

I hope this helps!

8 0

2 years ago

1 point) As reported in "Runner's World" magazine, the times of the finishers in the New York City 10 km run are normally distri

BigorU [14]

Answer:

(a) E (X) = 61 and SD (X) = 9

(b) E (Z) = 0 and SD (Z) = 1

Step-by-step explanation:

The time of the finishers in the New York City 10 km run are normally distributed with a mean,μ = 61 minutes and a standard deviation, σ = 9 minutes.

(a)

The random variable X is defined as the finishing time for the finishers.

Then the expected value of X is:

E (X) = 61 minutes

The variance of the random variable X is:

V (X) = (9 minutes)²

Then the standard deviation of the random variable X is:

SD (X) = 9 minutes

(b)

The random variable Z is the standardized form of the random variable X.

It is defined as: $Z=\frac{X-\mu}{\sigma}$

Compute the expected value of Z as follows:

$E(Z)=E[\frac{X-\mu}{\sigma}]\\=\frac{E(X)-\mu}{\sigma}\\=\frac{61-61}{9}\\=0$

The mean of Z is 0.

Compute the variance of Z as follows:

$V(Z)=V[\frac{X-\mu}{\sigma}]\\=\frac{V(X)+V(\mu)}{\sigma^{2}}\\=\frac{V(X)}{\sigma^{2}}\\=\frac{9^{2}}{9^{2}}\\=1$

The variance of Z is 1.

So the standard deviation is 1.

8 0

2 years ago