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2 years ago

The sum of 21 and n is equal to 43.23 please help me!!! I’ll give out 30 points!!!

Mathematics

2 answers:

Assoli18 [71]2 years ago

8 0

Equation: 21+n=43.23

Process to find n

21+n=43.23
Subtract 21 from both sides
n=22.23

12345 [234]2 years ago

3 0

Answer:

n=22.23

Step-by-step explanation:

21+n=43.23

-21 to both sides

0+n=22.23

n=22.23

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-13 can be classified as a whole number, interger, or rational number

beks73 [17]

Answer:

Step-by-step explanation:

Actually, -13 can be classified as any and all of these.

7 0

2 years ago

Read 2 more answers

4. If DF = 9x - 39, find EF. Help I’m so confused!!!!

topjm [15]

Answer:

EF =58

Step-by-step explanation:

from the illustration,

EF =DF - DE

given that DF =9x-39 , DE =47 EF=3x +10

substitute them into the formula,

3x +10=9x-39 - 47

solving for x

3x +10 =9x -86

grouping like terms

3x - 9x =-86 -10

-6x=-96

dividing through by -6

 $\frac{ - 6x}{ - 6} = \frac{ - 96}{ - 6}$ 

 $x = \frac{96}{6}$ 

 $x = 16$ 

but EF=3x+10

substitute x=16 into it to get the EF

EF= 3(16)+10

EF=48+10

EF=58

6 0

3 years ago

Read 2 more answers

A city is located at an elevation that is below sea level. what is a possible elevation for that city?

kvv77 [185]

Well Im just guessing but 1 1/2 foot

4 0

3 years ago

The life of a red bulb used in a traffic signal can be modeled using an exponential distribution with an average life of 24 mont

BartSMP [9]

Answer:

See steps below

Step-by-step explanation:

Let X be the random variable that measures the lifespan of a bulb.

If the random variable X is exponentially distributed and X has an average value of 24 month, then its probability density function is

$\bf f(x)=\frac{1}{24}e^{-x/24}\;(x\geq 0)$

and its cumulative distribution function (CDF) is

$\bf P(X\leq t)=\int_{0}^{t} f(x)dx=1-e^{-t/24}$

• What is probability that the red bulb will need to be replaced at the first inspection?

The probability that the bulb fails the first year is

$\bf P(X\leq 12)=1-e^{-12/24}=1-e^{-0.5}=0.39347$

• If the bulb is in good condition at the end of 18 months, what is the probability that the bulb will be in good condition at the end of 24 months?

Let A and B be the events,

A = “The bulb will last at least 24 months”

B = “The bulb will last at least 18 months”

We want to find P(A | B).

By definition P(A | B) = P(A∩B)P(B)

but B⊂A, so A∩B = B and

$\bf P(A | B) = P(B)P(B) = (P(B))^2$

We have

$\bf P(B)=P(X>18)=1-P(X\leq 18)=1-(1-e^{-18/24})=e^{-3/4}=0.47237$

hence,

$\bf P(A | B)=(P(B))^2=(0.47237)^2=0.22313$

• If the signal has six red bulbs, what is the probability that at least one of them needs replacement at the first inspection? Assume distribution of lifetime of each bulb is independent

If the distribution of lifetime of each bulb is independent, then we have here a binomial distribution of six trials with probability of “success” (one bulb needs replacement at the first inspection) p = 0.39347

Now the probability that exactly k bulbs need replacement is

$\bf \binom{6}{k}(0.39347)^k(1-0.39347)^{6-k}$

Probability that at least one of them needs replacement at the first inspection = 1- probability that none of them needs replacement at the first inspection.

This means that,

Probability that at least one of them needs replacement at the first inspection =

$\bf 1-\binom{6}{0}(0.39347)^0(1-0.39347)^{6}=1-(0.60653)^6=0.95021$

5 0

3 years ago

Jose wants to build a model of a 18-meter tall building. He will be using a scale

NeTakaya

You have to use a proportion.

3.5/18= 1.5/x
3.5x=27
x= 7.7 cm

8 0

2 years ago