If DE=52 and EF=21 Find DF

NikAS [45]

Justin rides 550 kilometers
or
justin rides 17 meter

6 0

3 years ago

A fitted multiple regression equation is Y = 28 + 5X1 - 4X2 + 7X3 + 2X4. When X1 increases 2 units and X2 increases 2 units as w

ad-work [718]

Answer:

A. Increase by 2

Step-by-step explanation:

Given that a fitted multiple regression equation is

$Y = 28 + 5X_1 -4X_2 + 7X_3 + 2X_4$

This is a multiple regression line with dependent variable y and independent variables x1, x2, x3 and x4

The coefficients of independent variables represent the slope.

In other words the coefficients represent the rate of change of y when xi is changed by 1 unit.

Given that x3 and x4 remain unchanged and x1 increases by 2 and x2 by 2 units

Since slope of x1 is 5, we find for one unit change in x1 we can have 5 units change in y

i.e. for 2 units change in x1, we expect 10 units change in Y

Similarly for 2 units change in x2, we expect -2(4) units change in Y

Put together we have

$10-8 =2$ change in y

Since positive 2, there is an increase by 2

A. Increase by 2

5 0

3 years ago

6A. make a box and whiskers plot for the following data label the extrems,quartites class score 40,30,75,82,90,25,80,46

Aliun [14]

<span> 25 30 35 40 46 60.5 75 80 81 82 90
min Q1 Q2/med Q3 max
81-35=46=Interquartile range

</span><span>Is that all the information you wanted?</span>

5 0

3 years ago

-4+w+17-2/3w=16 solve the equation

KatRina [158]

Answer:

w=9

Step-by-step explanation:

I hope this image is helpful to you.

.

4 0

3 years ago

A research study uses 800 men under the age of 55. Suppose that 30% carry a marker on the male chromosome that indicates an incr

ss7ja [257]

Answer:

a) There is a 12.11% probability that exactly 1 man has the marker.

b) There is a 85.07% probability that more than 1 has the marker.

Step-by-step explanation:

There are only two possible outcomes: Either the men has the chromosome, or he hasn't. So we use the binomial probability distribution.

Binomial probability

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

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

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 $\pi$ is the probability of X happening.

In this problem, we have that:

30% carry a marker on the male chromosome that indicates an increased risk for high blood pressure, so $\pi = 0.30$

(a) If 10 men are selected randomly and tested for the marker, what is the probability that exactly 1 man has the marker?

10 men, so $n = 10$

We want to find P(X = 1). So:

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

$P(X = 1) = C_{10,1}.(0.30)^{1}.(0.7)^{9} = 0.1211$

There is a 12.11% probability that exactly 1 man has the marker.

(b) If 10 men are selected randomly and tested for the marker, what is the probability that more than 1 has the marker?

That is $P(X > 1)$

We have that:

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

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

We also have that:

$P(X \leq 1) = P(X = 0) + P(X = 1)$

$P(X = 0) = C_{10,0}.(0.30)^{0}.(0.7)^{10} = 0.0282$

So

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0282 + 0.1211 = 0.1493$

Finally

$P(X > 1) = 1 - P(X \leq 1) = 1 - 0.1493 = 0.8507$

There is a 85.07% probability that more than 1 has the marker.

3 0

3 years ago