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3 years ago

Help please! Thank you very much.

Mathematics

1 answer:

valentina_108 [34]3 years ago

7 0

8 inches

It's asking about inches so first, you need to convert the feet into inches
So multiply 4*12=48
1/2*48= 24
2/3*48= 32

32-24= 8

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What is the value of the larger zero of the function: f(x) = 3x^2– 14x + 15?

AVprozaik [17]

Answer:

Step-by-step explanation:

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3 years ago

Use Euler’s Formula to find the missing number. Edges: 37 Faces: 25 Vertices: ?

matrenka [14]

Answer:

Step-by-step explanation:

Comment

The formula that relates edges faces and vertices is F + V = E + 2

Givens

Edges (E): 37

Faces (F) = 25

Vertices: x

Solution

25 + x = 37 + 2 Subtract 25 from both sides.

25-25 +x= 37 - 25 + 2 Combine

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2 years ago

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ruslelena [56]

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Step-by-step explanation:

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3 years ago

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, So

tatyana61 [14]

Answer:

The probability that at least 13 flights arrive late is 2.5196 $\times 10^{-6}$ .

Step-by-step explanation:

We are given that Southwest Air had the best rate with 80 % of its flights arriving on time.

A test is conducted by randomly selecting 18 Southwest flights and observing whether they arrive on time.

The above situation can be represented through binomial distribution;

$P(X = x) = \binom{n}{r}\times p^{r} \times (1-p)^{n-r} ; x = 0,1,2,3,.........$

where, n = number of trials (samples) taken = 18 Southwest flights

r = number of success = at least 13 flights arrive late

p = probability of success which in our question is probability that

flights arrive late, i.e. p = 1 - 0.80 = 20%

Let X = Number of flights that arrive late.

So, X ~ Binom(n = 18, p = 0.20)

Now, the probability that at least 13 flights arrive late is given by = P(X $\geq$ 13)

P(X $\geq$ 13) = P(X = 13) + P(X = 14) + P(X = 15) + P(X = 16) + P(X = 17) + P(X = 18)

= $\binom{18}{13}\times 0.20^{13} \times (1-0.20)^{18-13}+ \binom{18}{14}\times 0.20^{14} \times (1-0.20)^{18-14}+ \binom{18}{15}\times 0.20^{15} \times (1-0.20)^{18-15}+ \binom{18}{16}\times 0.20^{16} \times (1-0.20)^{18-16}+ \binom{18}{17}\times 0.20^{17} \times (1-0.20)^{18-17}+ \binom{18}{18}\times 0.20^{18} \times (1-0.20)^{18-18}$

= $\binom{18}{13}\times 0.20^{13} \times 0.80^{5}+ \binom{18}{14}\times 0.20^{14} \times 0.80^{4}+ \binom{18}{15}\times 0.20^{15} \times 0.80^{3}+ \binom{18}{16}\times 0.20^{16} \times 0.80^{2}+ \binom{18}{17}\times 0.20^{17} \times 0.80^{1}+ \binom{18}{18}\times 0.20^{18} \times 0.80^{0}$

= 2.5196 $\times 10^{-6}$ .

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Natali5045456 [20]

Answer: 10^-5 = 1/10^5

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1 year ago

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