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3 years ago

Use the probability distribution graph to answer the question. P(X≤a)=0.6 What is the value of a?

2 answers:

8 0

Answer: If P(x<a) = 0.6, then the value of a will be 0.25.

In this type of problem, it is safe to assume the normal distribution unless you are told something different.

In this probability, we have to find the z-score for a that gives a probability of 0.6. You will need a normal distribution chart to do this problem. Just find the probability that is closest to 0.6 and use that z-score.

quester [9]3 years ago

7 0

the answer will be 5

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Endpoint given the Midpoint: * The midpoint of RP is M(2, 4). If one of the end points is R(-1,7), find the coordinates of the o

marin [14]

Answer:

The coordinates of other point are: (5,1)

Step-by-step explanation:

Given coordinates are:

M(x,y) = (2,4)

R(x_1,x_2) = (-1,7)

We have to find the coordinates of other point (x2,y2)

The formula for mid-point is given by:

$M(x,y) = (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})$

Putting the values we get

$(2,4) = (\frac{-1+x_2}{2}, \frac{7+y_2}{2})$

Putting respective elements equal

$\frac{-1+x_2}{2} = 2\\-1+x_2 = 2*2\\-1+x_2 = 4\\x_2 = 4+1\\x_2 = 5\\And\\\frac{7+y_2}{2} = 4\\7+y_2 = 4*2\\7+y_2 = 8\\y_2 = 8-7\\y_2 = 1$

Hence,

The coordinates of other point are: (5,1)

8 0

3 years ago

Van someone help me with this

gizmo_the_mogwai [7]

Answer:

107

Step-by-step explanation:

5 0

3 years ago

Dominique decided to build his sons a fort in the backyard. He put a roof on the fort in the shape of a triangular prism.

VARVARA [1.3K]

Answer:

B. 216 ft

Step-by-step explanation:

4 0

2 years ago

Read 2 more answers

Suppose quantity s is a length and quantity t is a time. Suppose the quantities v and a are defined by v = ds/dt and a = dv/dt.

finlep [7]

Answer:

a) $v = \frac{[L]}{[T]} = LT^{-1}$

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

c) $\int v dt = s(t) = [L]=L$

d) $\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

e) $\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

Step-by-step explanation:

Let define some notation:

[L]= represent longitude , [T] =represent time

And we have defined:

s(t) a position function

$v = \frac{ds}{dt}$

$a= \frac{dv}{dt}$

Part a

If we do the dimensional analysis for v we got:

$v = \frac{[L]}{[T]} = LT^{-1}$

Part b

For the acceleration we can use the result obtained from part a and we got:

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

Part c

From definition if we do the integral of the velocity respect to t we got the position:

$\int v dt = s(t)$

And the dimensional analysis for the position is:

$\int v dt = s(t) = [L]=L$

Part d

The integral for the acceleration respect to the time is the velocity:

$\int a dt = v(t)$

And the dimensional analysis for the position is:

$\int a dt = v(t) = [L][T]^{-1}=LT^{-1}$

Part e

If we take the derivate respect to the acceleration and we want to find the dimensional analysis for this case we got:

$\frac{da}{dt}= \frac{[L][T]^{-2}}{T} = [L][T]^{-2} [T]^{-1} = LT^{-3}$

7 0

3 years ago

How do i solve 2.06 + [(3/5 - 1/2) × 2.5] ÷ 5

Diano4ka-milaya [45]

If you use pemdas you would get 2.11

3 0

3 years ago

Read 2 more answers