Please help! Nejcnrccn

(-4,7),(6,-1) what is the answer

Ronch [10]

(y+1)=-8/10 (x-6)
5y+5=-4x+24
5y=-4x+19

3 0

4 years ago

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Sorry about the writing, but could someone answer this by Friday? Thanks!

Paladinen [302]

The probability of a red marble is 25/250 or 1/10. The probability of a green marble is 50/250 or 1/5. The probability of a blue or orange marble is 175/250 or 7/10. Multiply the numerators and the denominators to get 7/500 probability (B)

4 0

3 years ago

The deciles of any distribution are the points that mark off the lowest 10% and the highest 10% of the distribution. the deciles

Levart [38]

(a) We need to figure out the deciles for a normal distribution curve in this part.

In order to do so, we need to find the z score corresponding to probability 0.10 and 0.90 on the normal distribution curve.

Upon looking at the normal distribution table, we can see that the probability P = 0.10 corresponds to a z score of -1.28 and probability P = 0.90 corresponds to a z score of 1.28.

Therefore, deciles are -1.28 and 1.28 for a normal distribution curve.

(b) In this part, we are given that mean height is 69.5 inches and standard deviation in height is 2.7 inches. Therefore, we can find the deciles using the z score formula as shown below:

$z=\frac{x-\mu}{\sigma }\Rightarrow \pm 1.28=\frac{x-69.5}{2.7}\\
x-69.5 = \pm 3.456\Rightarrow x=69.5\pm 3.456\Rightarrow \\
x=66.044,72.956$

Therefore, the required deciles are 66.044 and 72.956

7 0

3 years ago

The probability that a lab specimen contains high levels of contamination is 0.10. Five samples are checked, and the samples are

raketka [301]

Answer:

(a) 0.59049 (b) 0.32805 (c) 0.40951

Step-by-step explanation:

Let's define

$A_{i}$ : the lab specimen number i contains high levels of contamination for i = 1, 2, 3, 4, 5, so,

$P(A_{i})=0.1$ for i = 1, 2, 3, 4, 5

The complement for $A_{i}$ is given by

$A_{i}^{$c$}$ : the lab specimen number i does not contains high levels of contamination for i = 1, 2, 3, 4, 5, so

P( $A_{i}^{$c$}$ )=0.9 for i = 1, 2, 3, 4, 5

(a) The probability that none contain high levels of contamination is given by

P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )= $(0.9)^{5}$ =0.59049 because we have independent events.

(b) The probability that exactly one contains high levels of contamination is given by

P( $A_{1}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}$ ∩ $A_{5}^{$c$}$ )+P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}$ )=5×(0.1)× $(0.9)^{4}$ =0.32805

because we have independent events.

(c) The probability that at least one contains high levels of contamination is

P( $A_{1}$ ∪ $A_{2}$ ∪ $A_{3}$ ∪ $A_{4}$ ∪ $A_{5}$ )=1-P( $A_{1}^{$c$}$ ∩ $A_{2}^{$c$}$ ∩ $A_{3}^{$c$}$ ∩ $A_{4}^{$c$}$ ∩ $A_{5}^{$c$}$ )=1-0.59049=0.40951

6 0

3 years ago

why is it that all people have to do to get help is just say "HHEEELLLLPPPPP MMMMEEEEEE PPPPLLLLZZZZZZZ!!!!!!!!!!!!!!"

mart [117]

Answer:

Lol that's what I do and it helps lollll

3 0

3 years ago

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