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3 years ago

What is 28 equal to 14 plus 7w

Mathematics

2 answers:

Kisachek [45]3 years ago

8 0

Answer:

2=w

Step-by-step explanation:

28 = 14 + 7w

-14 -14

14 = 7w

14/7 = 2

7w/7 = w

zmey [24]3 years ago

4 0

Hello there.

The answer to this problem would be 2=w. To solve, simply subtract 14 from 28. The answer would be 2. We do this since we cannot add a number to variable, so we just add or subtract the number to number. Finally, divide both sides by 7 since you want to get rid of 7w. Then your answer should be 2=w.

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The perimeter of the rectangular pool is 80 feet.The length of the pool is 10 feet more than twice of the width.Find the length

AveGali [126]

Width= 10
Length= 30

Explanation: photo

7 0

2 years ago

Because all airline passengers do not show up for their reserved seat, an airline sells 125 tickets for a ﬂight that holds only

bekas [8.4K]

Answer: a) 0.9961, b) 0.9886

Step-by-step explanation:

Since we have given that

Probability that does not show up = 0.10

Probability that show up = 0.90

Here, we use "Binomial distribution":

n = 125 and p = 0.90

Number of passengers that hold in a flight = 120

a) What is the probability that every passenger who shows up can take the ﬂight?

$P(X\leq 120)=\sum_{x=0}^{120}^{125}C_x(0.90)^x(0.10)^{125-x}=0.9961$

(b) What is the probability that the ﬂight departs with empty seats?

$P(X\leq 119)=\sum _{x=0}^{119}^{125}C_x(0.90)^x)(0.10)^{125-x}=0.9886$

Hence, a) 0.9961, b) 0.9886

3 0

3 years ago

A major cab company in Chicago has computed its mean fare from O'Hare Airport to the Drake Hotel to be $28.75 , with a standard

bulgar [2K]

Answer:

Following are the solution to the given choices:

Step-by-step explanation:

Using chebyshev's theorem :

$\to P(|(x-\mu)|\leq k \sigma)\geq 1- \frac{1}{k^2}\\\\here \\ \to \mu=28.75\\\\ \to \sigma=4.44$

In point a)

$\to |(x-\mu)|= |20.17-28.75|=8.58 and \sigma=4.44 \\\\so, \\k= \frac{ |(x-\mu)| }{\sigma}$

$= \frac{8.58}{4.44} \\\\ =1.9 \\\\ =2$

$value= (1- \frac{1}{k^2}) \times 100 \% =75\%$

In point b)

$\to (1- \frac{1}{k^2}) \times 100 \% =84\% \\\\\to (1- \frac{1}{k^2})=0.84 \\\\\to \frac{1}{k^2} =0.16 \\\\\to \frac{1}{k}=0.4\\\\ \to k=2.5 \\\\\to |(x-\mu)| \leq k \sigma \\\\= |(x-\mu)|\leq 2.5 \times 4.44 \\\\ = |(x-\mu)|\leq 1.11 \\\\ = (28.75\pm 11.1) \\\\\to \text{fares lies between}(17.65,39.85)$