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2 years ago

Matt made the model below to help him solve his math problem.

Mathematics

1 answer:

bonufazy [111]2 years ago

6 0

Answer:

6X5

Step-by-step explanation:

There are 6 columns and 5 rows. Therefore it should be 6X5.

I hope this helped, have a great day :D

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Estimate 19 precent of 41

Burka [1]

19% turns into 20%

20%*41=8.2

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2 years ago

A hotel house keeper cleaned 1/3 on

Aleksandr-060686 [28]

Answer:

dunno

Step-by-step explanation:

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3 0

2 years ago

Solve for m. 774 = m − 15

lyudmila [28]

Answer:

$m = 789$

Step-by-step explanation:

$774 = m - 15 \\ 774 + 15 = m \\ 789 = m$

hope this helps

brainliest appreciated

good luck! have a nice day!

6 0

3 years ago

Read 2 more answers

Four polynomials are shown below:

Tatiana [17]

The appropriate choice is
Polynomial B. -3x⁴ +5x³ +5

_____
C and D are not trinomials. (They have 4 terms, not 3.) A is 5th degree, not 4th.

8 0

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 $\times 10^{-6}$ .

7 0

3 years ago