Whats the value of x ?

Which scatter plot shows a definite non-linear relationship between x and y?

Arturiano [62]

Maybe c or A not completely sure try looking for the one you think it is and try reading it again

6 0

3 years ago

Consider the arithmetic sequence 13, 24, 35, ....

Nostrana [21]

Answer: 27

Step-by-step explanation:

The given sequence : 13, 24, 35, ....

Here, first term : a= 13

Common difference: d=24-13 =11

The nth term of arithmetic sequence is given by :-

$a_n=a+(n-1)d$

It is given that the nth term is 299, so put $a_n=299, a=13 ; d=11$

$299=13+(n-1)11$

Subtract 13 from both sides , we get

$286=(n-1)11$

Divide both sides by 11, we get

$26=n-1$

Add 1 on both sides , we get

$n=27$

Hence, the value of n= 27

4 0

3 years ago

At an ocean-side nuclear power plant, seawater is used as part of the cooling system. This raises the temperature of the water t

grandymaker [24]

Answer:

(a1) The probability that temperature increase will be less than 20°C is 0.667.

(a2) The probability that temperature increase will be between 20°C and 22°C is 0.133.

(b) The probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c) The expected value of the temperature increase is 17.5°C.

Step-by-step explanation:

Let X = temperature increase.

The random variable X follows a continuous Uniform distribution, distributed over the range [10°C, 25°C].

The probability density function of X is:

$f(X)=\left \{ {{\frac{1}{25-10}=\frac{1}{15};\ x\in [10, 25]} \atop {0;\ otherwise}} \right.$

(a1)

Compute the probability that temperature increase will be less than 20°C as follows:

$P(X$

Thus, the probability that temperature increase will be less than 20°C is 0.667.

(a2)

Compute the probability that temperature increase will be between 20°C and 22°C as follows:

$P(20$

Thus, the probability that temperature increase will be between 20°C and 22°C is 0.133.

(b)

Compute the probability that at any point of time the temperature increase is potentially dangerous as follows:

$P(X>18)=\int\limits^{25}_{18}{\frac{1}{15}}\, dx\\=\frac{1}{15}\int\limits^{25}_{18}{dx}\,\\=\frac{1}{15}[x]^{25}_{18}=\frac{1}{15}[25-18]=\frac{7}{15}\\=0.467$

Thus, the probability that at any point of time the temperature increase is potentially dangerous is 0.467.

(c)

Compute the expected value of the uniform random variable X as follows:

$E(X)=\frac{1}{2}[10+25]=\frac{35}{2}=17.5$

Thus, the expected value of the temperature increase is 17.5°C.

7 0

3 years ago

What makes b in the equation true 3/8 X 8/b

Gala2k [10]

$\frac{3}{8} * \frac{8}{b} = \frac{3}{1} * \frac{1}{b}$ (I crossed simplify)

Next cross multiply;

3b = 1

b = 1/3

To check if the statement is true just substitute the value of b in the equation;

3b = 1

3(1/3) = 1

1 = 1

8 0

3 years ago

What is the slope of the line? A. 2 B. -2 C. 1/2 D. -1/2

iogann1982 [59]

slope m = - 2 → B

calculate m using the ' gradient formula '

m = (y₂ - y₁)/(x₂ - x₁)

choosing 2 'clear ' coordinates from the graph of the line

(x₁, y₁) = (0, 7) and (x₂, y₂) = (7, - 5 $\frac{(- 5 - 7)}{(6 - 0)}$ ) = $\frac{- 12}{6}$ = - 2

m =

3 0

3 years ago