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3 years ago

Can someone solve these 1.) 4 + c =6

Mathematics

1 answer:

Setler79 [48]3 years ago

8 0

Answer:

1. c=1

2. x=20

3. c=3

4. x=40

Step-by-step explanation:

Hope this helps, lemme know if you need this explained (:

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Please help I’m failing math

jasenka [17]

Answer:

3x+3

Step-by-step explanation:

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3 years ago

A set of data already has the values 11, 14, 23, and 16. What value would have to be added to the set for the mean of the five n

KonstantinChe [14]

1) Take 18 and multiply it by 5 = 90
2) Then subtract 26 = 64
3) Then subtract 18 = 46
4) Then subtract 12 = 34
5) Then subtract 8 = 26

The answer is 26.

5 0

3 years ago

What is the percent of change from 14 inches 26 inches

PolarNik [594]

To find the percent of anything, you divide what you have from the total (or the smaller number over the bigger number). In this case, you will do 14/26.
14/26=0.53846
I'm going to round to the hundredths: 0.54
Move the decimal place two times to the right: 54
The percent change from 14 inches to 26 inches is 54%

4 0

3 years ago

What is the product of (n - 8)(n + 2)?

Anuta_ua [19.1K]

Answer:

n² - 6n - 16

Step-by-step explanation:

As per the question,

We have been provided the equation

(n-8)(n+2)

To find the product, we have to multiply the both term,

(n - 8) (n + 2)

= n (n + 2) - 8 (n + 2)

= (n² + 2n) - (8n + 16)

= n² + 2n - 8n - 16n

= n² - 6n - 16

Hence, the required answer of the product of (n-8)(n+2) = n² - 6n - 16.

5 0

3 years ago

Read 2 more answers

Suiting at 6 a.m., cars, buses, and motorcycles arrive at a highway loll booth according to independent Poisson processes. Cars

dem82 [27]

Answer:

Step-by-step explanation:

From the information given:

the rate of the cars = $\dfrac{1}{5} \ car / min = 0.2 \ car /min$

the rate of the buses = $\dfrac{1}{10} \ bus / min = 0.1 \ bus /min$

the rate of motorcycle = $\dfrac{1}{30} \ motorcycle / min = 0.0333 \ motorcycle /min$

The probability of any event at a given time t can be expressed as:

$P(event \ (x) \ in \ time \ (t)\ min) = \dfrac{e^{-rate \times t}\times (rate \times t)^x}{x!}$

∴

(a)

$P(2 \ car \ in \ 20 \ min) = \dfrac{e^{-0.20\times 20}\times (0.2 \times 20)^2}{2!}$

$P(2 \ car \ in \ 20 \ min) =0.1465$

$P ( 1 \ motorcycle \ in \ 20 \ min) = \dfrac{e^{-0.0333\times 20}\times (0.0333 \times 20)^1}{1!}$

$P ( 1 \ motorcycle \ in \ 20 \ min) = 0.3422$

$P ( 0 \ buses \ in \ 20 \ min) = \dfrac{e^{-0.1\times 20}\times (0.1 \times 20)^0}{0!}$

$P ( 0 \ buses \ in \ 20 \ min) = 0.1353$

Thus;

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.1465 × 0.3422 × 0.1353

P(exactly 2 cars, 1 motorcycle in 20 minutes) = 0.0068

(b)

the rate of the total vehicles = 0.2 + 0.1 + 0.0333 = 0.3333

the rate of vehicles with exact change = rate of total vehicles × P(exact change)

$= 0.3333 \times \dfrac{1}{4}$

= 0.0833

∴

$P(zero \ exact \ change \ in \ 10 minutes) = \dfrac{e^{-0.0833\times 10}\times (0.0833 \times 10)^0}{0!}$

P(zero exact change in 10 minutes) = 0.4347

The probability of the 7th motorcycle after the arrival of the third motorcycle is:

$P( 4 \ motorcyles \ in \ 45 \ minutes) =\dfrac{e^{-0.0333\times 45}\times (0.0333 \times 45)^4}{4!}$

$P( 4 \ motorcyles \ in \ 45 \ minutes) =0.0469$

Thus; the probability of the 7th motorcycle after the arrival of the third one is = 0.0469

P(at least one other vehicle arrives between 3rd and 4th car arrival)

= 1 - P(no other vehicle arrives between 3rd and 4th car arrival)

The 3rd car arrives at 15 minutes

The 4th car arrives at 20 minutes

The interval between the two = 5 minutes

<u>For Bus:</u>

P(no other vehicle other vehicle arrives within 5 minutes is)

= $\dfrac{6}{12} = 0.5$

<u>For motorcycle:</u>

$= \dfrac{2 }{12} = \dfrac{1 }{6}$

∴

The required probability = $1 - \Bigg ( \dfrac{e^{-0.5 \times 0.5^0}}{0!} \times \dfrac{e^{-1/6}\times (1/6)^0}{0!} \Bigg)$

= 1- 0.5134

= 0.4866

6 0

3 years ago