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aleksley [76]
3 years ago
7

What percentage of the data values falls between the values of 3 and 24 in the data set shown?

Mathematics
2 answers:
RSB [31]3 years ago
6 0
A.25% yeah bc yeah yeah yeah yeah
Nitella [24]3 years ago
3 0
I would say 25% because it’s not half of that 3 is not half of 24 so it can’t be 50 so 25 would be your best choose
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* please help 20 points !! *<br> which statement can be proved if you are given that SK≌LR?
SOVA2 [1]
I believe it would be D) ST ≈ TR
6 0
3 years ago
Read 2 more answers
The  area of triangle formed by points of intersection of parabola y=a(x+5)(x−1) with the coordinate axes is 12. Find a if it is
Pie

We know that the points at which the parabola intersects the x axis are

(-5,0) and (1,0)

So the extent between these two points would be the base of the triangle

lets find the length of the base using the distance formula

\sqrt{[(-5-1)^{2}+(0-0)^{2} ]}

the base b=6

We will get the height of the triangle when we put x=0 in the equation

y=a(0+5)(0-1)

y=-5a

so height = -5a (we take +5a since it is the height)

We know that the area of the triangle =

\frac{1}{2} × 6 × (5a) = 12

15a=12

a= \frac{4}{5}


7 0
3 years ago
(3,5),slope -2<br><br> Show work
vova2212 [387]

Answer:

y=-2x+11

Step-by-step explanation:

equation of a line

y-b=m(x-a)

m=-2

a=3

b=5

sub into the equation

y-5= -2(x-3) multiply out the brackets and a negative multiply by negative equals positive

y-5=-2x+6

y=-2x+11

or

2x+y-11=0

8 0
3 years ago
In the library on a university campus, there is a sign in the elevator that indicates a limit of 16 persons. Furthermore, there
Vedmedyk [2.9K]

Answer:

a) 151lb.

b) 6.25 lb

Step-by-step explanation:

The Central Limit Theorem estabilishes that, for a random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation \frac{\sigma}{\sqrt{n}}.

In this problem, we have that:

\mu = 151, \sigma = 25, n = 16

So

a) The expected value of the sample mean of the weights is 151 lb.

(b) What is the standard deviation of the sampling distribution of the sample mean weight?

This is s = \frac{\sigma}{\sqrt{n}} = \frac{25}{\sqrt{16}} = 6.25

8 0
3 years ago
Pls answer quickly :)
baherus [9]

Answer:

median=mode

Step-by-step explanation:

7 0
3 years ago
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