4/7 - (-4/7) PLZ HELP!

Point Q(-2, 8) is translated using the rule (x + 1, y - 2).

insens350 [35]

The coordinate notation is (x,y)---- >(x+1,y-2)

Q(-2, 8)

only y coordinate is (y)--------------- > ( y-2)
therefore
(8)------------------------------------ >(8-2)=(6)
the y coordinate of Q' is 6

7 0

3 years ago

Please help and please slove and explain im not getting this

OLga [1]

Answer:

x-intercept= (3⅓, 0)

y-intercept= (0, 4²⁄₇)

Please see the attached picture for the graph.

Step-by-step explanation:

-9x -7y= -30

Let's simplify the equation by dividing both sides by -1.

9x +7y= 30

x- intercept occurs at y= 0.

When y= 0,

9x +7(0)= 30

9x= 30

x= 30 ÷9

$x = 3 \frac{1}{3}$

Thus, x- intercept occurs at (3⅓, 0).

y-intercept occurs at x= 0.

When x= 0,

9(0) +7y= 30

7y= 30

y= 30 ÷7

$y = 4 \frac{2}{7}$

Thus, y- intercept occurs at (0, 4²⁄₇).

_____

To graph the equation, draw the x and y axis on a graph paper. Use an appropriate scale to divide the line into equal parts. Next, plot the points (3⅓, 0) and (0, 4²⁄₇). Then, join the two points with a straight line.

Notes:

x- intercept is the point at which the graph cuts through the x- axis. In this case, your x- axis is the horizontal line that runs from left to right of your graph paper. In order for a point to be on this horizontal line, look at the y- axis and notice that it sits at y= 0. The same reason applies for why the y- intercept occurs at x= 0. This has to do with the two axis cutting each other at the point (0,0), resulting in the x and y coordinates of 0 for the y and x intercepts respectively.

Simplifying the equation in the first step is not necessary, but it is a good practice and might reduce carelessness.

5 0

2 years ago

Among persons donating blood to a clinic, 85% have Rh+ blood (that is, the Rhesus factor is present in their blood.) Six people

Leona [35]

Answer:

a) There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) There is a 22.36% probability that at most four of the six have Rh+ blood.

c) There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

Step-by-step explanation:

For each person donating blood, there are only two possible outcomes. Either they have Rh+ blood, or they do not. This means that we use the binomial probability distribution to solve this problem.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

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

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

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

And p is the probability of X happening.

In this problem we have that:

$p = 0.85, n = 6$ .

a) fine the probability that at least one of the five does not have the Rh factor.

Either all six have the factor, or at least one of them do not. The sum of the probabilities of these events is decimal 1. So:

$P(X < 6) + P(X = 6) = 1$

$P(X < 6) = 1 - P(X = 6)$

In which:

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

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

So

$P(X < 6) = 1 - P(X = 6) = 1 - 0.3771 = 0.6229$

There is a 62.29% probability that at least one of the five does not have the Rh factor.

b) find the probability that at most four of the six have Rh+ blood.

Either more than four have Rh+ blood, or at most four have. So

$P(X \leq 4) + P(X > 4) = 1$

$P(X \leq 4) = 1 - P(X > 4)$

In which

$P(X > 4) = P(X = 5) + P(X = 6)$

$P(X = 5) = C_{6,5}.(0.85)^{5}.(0.15)^{1} = 0.3993$

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

$P(X > 4) = P(X = 5) + P(X = 6) = 0.3993 + 0.3771 = 0.7764$

$P(X \leq 4) = 1 - P(X > 4) = 1 - 0.7764 = 0.2236$

There is a 22.36% probability that at most four of the six have Rh+ blood.

c) The clinic needs six Rh+ donors on a certain day. How many people must donate blood to have the probability of obtaining blood from at least six Rh+ donors over 0.95?

With 6 donors:

$P(X = 6) = C_{6,6}.(0.85)^{6}.(0.15)^{0} = 0.3771$

37.71% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 7 donors:

$P(X = 6) = C_{7,6}.(0.85)^{6}.(0.15)^{1} = 0.3960$

0.3771 + 0.3960 = 0.7764 = 77.64% probability of obtaining blood from at least six Rh+ donors over 0.95.

With 8 donors

$P(X = 6) = C_{8,6}.(0.85)^{6}.(0.15)^{2} = 0.2376$

0.3771 + 0.3960 + 0.2376 = 1.01 = 101% probability of obtaining blood from at least six Rh+ donors over 0.95.

There need to be at least 8 people to have the probability of obtaining blood from at least six Rh+ donors over 0.95.

5 0

2 years ago

Write an algebraic expression for the word expression 11 more than e

solong [7]

Answer: e + 11

Step-by-step explanation:

6 0

2 years ago

Practice 7) fill in the blanks.

DiKsa [7]

(h,k) is the vertex

b/2a is the line of symmetry

a > 0 parabola opens upward, vertex is a minimum

a < 0 parabola opens downward, vertex is a maximum

My first answer was deleted lolol

7 0

2 years ago