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Ludmilka [50]
3 years ago
9

A net has a rectangle at the center and 4 triangles on the sides. The rectangle has a length of 6 inches and height of 4 inches.

2 triangles have a base of 6 inches and height of 10 inches. 2 triangles have a base of 4 inches and height of 10.4 inches. What are the areas of the faces of the rectangular pyramid? Select all that apply. 12 in.2 24 in.2 30 in.2 20.8 in.2 60 in.2 61.2 in.2 plz help me
Mathematics
2 answers:
german3 years ago
8 0

Answer:

b,c,d i hoped this helped

Step-by-step explanation:

Question

A net has a rectangle at the center and 4 triangles on the sides. The rectangle has a length of 6 inches and height of 4 inches. 2 triangles have a base of 6 inches and height of 10 inches. 2 triangles have a base of 4 inches and height of 10.4 inches. What are the areas of the faces of the rectangular pyramid? Select all that apply. 12 in.2 24 in.2 30 in.2 20.8 in.2 60 in.2 61.2 in.2  

Oduvanchick [21]3 years ago
6 0

Answer:

its b c and d

Step-by-step explanation:

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The Hawaiian language has 12 letters: five vowels and seven consonants. Each of the 12 Hawaiian letters are written on a slip of
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12 total letters...5 vowels and 7 consonants.

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after replacing ...
probability of 2nd letter being a vowel is 5/12

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3 years ago
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The number of text messages sent daily by a student is a poisson random variable with parameter λ=5 .in a class with 20 independ
Rudik [331]
This problem is a combination of the Poisson distribution and binomial distribution.

First, we need to find the probability of a single student sending less than 6 messages in a day, i.e.
P(X<6)=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)
=0.006738+0.033690+0.084224+0.140374+0.175467+0.175467
= 0.615961

For ALL 20 students to send less than 6 messages, the probability is
P=C(20,20)*0.615961^20*(1-0.615961)^0
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7 0
4 years ago
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!
Anettt [7]

Answer:

Choice C: approximately 121 green beans will be 13 centimeters or shorter.

Step-by-step explanation:

What's the probability that a green bean from this sale is shorter than 13 centimeters?

Let the length of a green bean be X centimeters.

X follows a normal distribution with

  • mean \mu = 11.2 and
  • standard deviation \sigma = 2.1.

In other words,

X\sim \text{N}(11.2, 2.1^{2}),

and the probability in question is X \le 13.

Z-score table approach:

Find the z-score of this measurement:

\displaystyle z= \frac{x-\mu}{\sigma} = \frac{13-11.2}{2.1} = 0.857143. Closest to 0.86.

Look up the z-score in a table. Keep in mind that entries on a typical z-score table gives the probability of the left tail, which is the chance that Z will be less than or equal to the z-score in question. (In case the question is asking for the probability that Z is greater than the z-score, subtract the value from table from 1.)

P(X\le 13) = P(Z \le 0.857143) \approx 0.8051.

"Technology" Approach

Depending on the manufacturer, the steps generally include:

  • Locate the cumulative probability function (cdf) for normal distributions.
  • Enter the lower and upper bound. The lower bound shall be a very negative number such as -10^{9}. For the upper bound, enter 13
  • Enter the mean and standard deviation (or variance if required).
  • Evaluate.

For example, on a Texas Instruments TI-84, evaluating \text{normalcdf})(-1\text{E}99,\;13,\;11.2,\;2.1 ) gives 0.804317.

As a result,

P(X\le 13) = 0.804317.

Number of green beans that are shorter than 13 centimeters:

Assume that the length of green beans for sale are independent of each other. The probability that each green bean is shorter than 13 centimeters is constant. As a result, the number of green beans out of 150 that are shorter than 13 centimeters follow a binomial distribution.

  • Number of trials n: 150.
  • Probability of success p: 0.804317.

Let Y be the number of green beans out of this 150 that are shorter than 13 centimeters. Y\sim\text{B}(150,0.804317).

The expected value of a binomial random variable is the product of the number of trials and the probability of success on each trial. In other words,

E(Y) = n\cdot p = 150 \times 0.804317 = 120.648\approx 121

The expected number of green beans out of this 150 that are shorter than 13 centimeters will thus be approximately 121.

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