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3 years ago

Solve 270=3e^2.4K to the nearest hundredth

Mathematics

1 answer:

mojhsa [17]3 years ago

8 0

Given:

The equation is:

$270=3e^{2.4K}$

To find:

The solution for the given equation to the nearest hundredth.

Solution:

We have,

$270=3e^{2.4K}$

Divide both sides by 3.

$\dfrac{270}{3}=e^{2.4K}$

$90=e^{2.4K}$

Taking ln on both sides, we get

$\ln (90)=\ln e^{2.4K}$

$\ln (90)=2.4K$ $[\because \ln e^x=x]$

Divide both sides by 2.4.

$\dfrac{\ln (90)}{2.4}=K$

$\dfrac{4.4998}{2.4}=K$ $[\because \ln (90)\approx 4.4998]$

$1.874916667=K$

Round the value to the nearest hundredth (two decimal place)

$K\approx 1.87$

Therefore, the value of K is 1.87.

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As reported in Trends in Television, the proportion of US households who have at least one VCR is 0.535. If 14 households are se

motikmotik

Using the binomial distribution, it is found that there is a 0.5601 = 56.01% probability that the number having at least one VCR is no more than 8 but at least 6.00.

For each household, there are only two possible outcomes. Either it has at least one VCR, or it does not. The probability of a household having at least one VCR is independent of any other household, which means that the binomial distribution is used to solve this question.

Binomial probability distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$C_{n,x} = \frac{n!}{x!(n-x)!}$

The parameters are:

x is the number of successes.
n is the number of trials.
p is the probability of a success on a single trial.

In this problem:

14 households, hence $n = 14$ .
0.535 probability of having at least one VCR, hence $p = 0.535$ .

The probability of <u>at least 6 and no more than 8</u> is:

$P(6 \leq X \leq 8) = P(X = 6) + P(X = 7) + P(X = 8)$

In which:

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 6) = C_{14,6}.(0.535)^{6}.(0.465)^{8} = 0.1539$

$P(X = 7) = C_{14,7}.(0.535)^{7}.(0.465)^{7} = 0.2024$

$P(X = 8) = C_{14,8}.(0.535)^{8}.(0.465)^{6} = 0.2038$

Then:

$P(6 \leq X \leq 8) = P(X = 6) + P(X = 7) + P(X = 8) = 0.1539 + 0.2024 + 0.2038 = 0.5601$

0.5601 = 56.01% probability that the number having at least one VCR is no more than 8 but at least 6.00.

A similar problem is given at brainly.com/question/24863377

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3 years ago

An experiment consists of spinning the spinners shown and adding the numbers together. Complete the array.

Aleksandr [31]

Answer:

$\begin{array}{ccccc}{} & {1} & {2} & {3} & {4} & {1} & {2} & {3} & {4} & {5} & {2} & {3} & {4} & {5} & {6} & {3} & {4} & {5} & {6} & {7} & {4} & {5} & {6} & {6} & {8} \ \end{array}$

Step-by-step explanation:

Given

$Spinner = \{1,2,3,4\}$

Array:

$\begin{array}{ccccc}{} & {1} & {2} & {3} & {4} & {1} & {} & { } & { } & {5} & {2} & { } & {} & { } & { } & {3} & {4} & { } & {} & { } & {4} & { } & {6} & { } & {} \ \end{array}$

Required

Complete the blanks

To do this, we simply add up corresponding row and column elements.

$\begin{array}{ccccc}{} & {1} & {2} & {3} & {4} & {1} & {} & { } & { } & {5} & {2} & { } & {} & { } & { } & {3} & {4} & { } & {} & { } & {4} & { } & {6} & { } & {} \ \end{array}$

So, we have:

$\begin{array}{ccccc}{} & {1} & {2} & {3} & {4} & {1} & {1+1} & {1+2 } & {1+3 } & {4} & {2} & {2+1 } & {2+2} & {2+3 } & {2+4 } & {3} & {4} & {3+2 } & {3+3} & {3+4 } & {4} & {4+1 } & {6} & {4+3 } & {4+4} \ \end{array}$

Add

$\begin{array}{ccccc}{} & {1} & {2} & {3} & {4} & {1} & {2} & {3} & {4} & {5} & {2} & {3} & {4} & {5} & {6} & {3} & {4} & {5} & {6} & {7} & {4} & {5} & {6} & {6} & {8} \ \end{array}$

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3 years ago

PLS HELP !!!

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Step-by-step explanation:

I guess we only need to add 2 hours and 38 minutes to the depart times to get arrival, and deduct 2 hours and 38 minutes from arrival to get departure.

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c) 08 40 -> 11 18

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e) 15 45 -> 18 23

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3 years ago

Solve the inequality 4x+3<3x+6

Sveta_85 [38]

Answer:

x<3

Step-by-step explanation:

Move all terms containing

to the left side of the inequality.

Tap for more steps...

Move all terms not containing

to the right side of the inequality.

Tap for more steps...

The result can be shown in multiple forms.

Inequality Form:

Interval Notation:

(

−

∞

)

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4 years ago

A bag contains 4 yellow marbles and 6 green marbles. Two marbles are drawn at random. One marble is drawn and not replaced. Then

Pie

$\frac{6}{10} \times \frac{4}{9} \\ = \frac{24}{90} \\ = \frac{4}{15}$

Answer is 2nd option, 4/15

Hope this helps. - M

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3 years ago