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3 years ago

What is the value of 8(2.3+x) when x=7

2 answers:

3 0

Just substitute the value of x = 7
= 8(2.3 x)
= 8(2.3 * 7)
= 8(16.1)
= 128.8

In short, Your Final Answer would be 128.8

Hope it help

alukav5142 [94]3 years ago

3 0

Hello!

The first thing you want to do it plug in the numbers.

8(2.3+7) = 74.4

Hope this helps!

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Please help me on number 5! asap please!

kirill115 [55]

The answer is 13 you subtract 25 and 12

3 0

3 years ago

There is a 20% chance of thunderstorms tomorrow. Describe the likelihood of the event.

Pachacha [2.7K]

Answer:

Step-by-step explanation:

20% = 1/5 or 0.2

There is a 1/5 chance of a thunderstorm.

There really is no clear answer. I would try 0.2

8 0

3 years ago

Read 2 more answers

Suppose that in a random sample of size 200, standard deviation of the sampling distribution of the sample mean 0.08. researcher

sertanlavr [38]

Answer:

400

Step-by-step explanation:

The computation of the sample size needed is shown below:

Since at the sample size of 200 there is a standard deviation of 0.08

But when the standard deviation is 0.04 so the sample size is 400

As sample standard deviation would be inversely proportional to the square root of the sample size

7 0

3 years ago

a vat of milk has spilled on a tile floor. the milk flow can be expressed with the function r(t) = 4t, where t represents time i

salantis [7]

4t=r
a=pir^2
sub 4t for r
a=pi(4t)^2
a=pi16t^2
a(t)=16pi(t^2)

A. a(t)=16pi(t^2)

B. sub 4 for t
a(4)=16pi4^2
a(4)=16pi16
a(4)=16*16*3.14
a(4)=803.84 square units

A. a(t)=16pi(t^2)
B. 803.84 square units

4 0

3 years ago

Read 2 more answers

The diameter of a particle of contamination (in micrometers) is modeled with the probability density function f(x)= 2/x^3 for x

natulia [17]

Answer:

a) 0.96

b) 0.016

c) 0.018

d) 0.982

e) x = 2

Step-by-step explanation:

We are given with the Probability density function f(x)= 2/x^3 where x > 1.

<em>Firstly we will calculate the general probability that of P(a < X < b) </em>

P(a < X < b) = $\int_{a}^{b} \frac{2}{x^{3}} dx$ = $2\int_{a}^{b} x^{-3} dx$

= $2[ \frac{x^{-3+1} }{-3+1}]^{b}_a dx$ { Because $\int_{a}^{b} x^{n} dx = [ \frac{x^{n+1} }{n+1}]^{b}_a$ }

= $2[ \frac{x^{-2} }{-2}]^{b}_a$ = $\frac{2}{-2} [ x^{-2} ]^{b}_a$

= $-1 [ b^{-2} - a^{-2} ]$ = $\frac{1}{a^{2} } - \frac{1}{b^{2} }$

a) Now P(X < 5) = P(1 < X < 5) {because x > 1 }

Comparing with general probability we get,

P(1 < X < 5) = $\frac{1}{1^{2} } - \frac{1}{5^{2} }$ = $1 - \frac{1}{25}$ = 0.96 .

b) P(X > 8) = P(8 < X < ∞) = 1/ $8^{2}$ - 1/∞ = 1/64 - 0 = 0.016

c) P(6 < X < 10) = $\frac{1}{6^{2} } - \frac{1}{10^{2} }$ = $\frac{1}{36} - \frac{1}{100 }$ = 0.018 .

d) P(x < 6 or X > 10) = P(1 < X < 6) + P(10 < X < ∞)

= $(\frac{1}{1^{2} } - \frac{1}{6^{2} })$ + (1/ $10^{2}$ - 1/∞) = 1 - 1/36 + 1/100 + 0 = 0.982

e) We have to find x such that P(X < x) = 0.75 ;

⇒ P(1 < X < x) = 0.75

⇒ $\frac{1}{1^{2} } - \frac{1}{x^{2} }$ = 0.75

⇒ $\frac{1} {x^{2} }$ = 1 - 0.75 = 0.25

⇒ $x^{2}$ = $\frac{1}{0.25}$ ⇒ $x^{2}$ = 4 ⇒ x = $2$

Therefore, value of x such that P(X < x) = 0.75 is 2.

8 0

3 years ago