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katen-ka-za [31]
2 years ago
14

Which of the following is a valid probability distribution? A 2-column table labeled Probability Distribution A has 4 rows. The

first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.42, 0.38, 0.13, 0.07. A 2-column table labeled Probability Distribution B has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.27, 0.28, 0.26, 0.27. A 2-column table labeled Probability Distribution C has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.16, 0.39, 0.18, 0.17. A 2-column table labeled Probability Distribution D has 4 rows. The first column is labeled x with entries 1, 2, 3, 4. The second column is labeled P (x) with entries 0.63, 0.12, 0.14, 0.13.
Mathematics
1 answer:
dsp732 years ago
8 0

Answer:

I believe it is A

Step-by-step explanation:

all variables should add up to 1.00 and all in the first answer add up to 1.00

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In a golf tournament, Sam played 4 rounds and was within 2 strokes of par for all 4 rounds of the tournament. If par is 72 on th
In-s [12.5K]

Given:

|x-4(72)|=2

To find:

The highest and lowest scores Sam could have made in the tournament.

Solution:

We have,

|x-4(72)|=2

|x-288|=2

It can be written as

x-288=\pm 2

Add 288 on both sides.

x=288\pm 2

x=288-2 and x=288+2

x=286 and x=290

Therefore, the highest and lowest scores Sam could have made in the tournament are 290 and 286 respectively.

8 0
1 year ago
A conical pile of road salt has a diameter of 112 feet and a slant height of 65 feet. After a storm, the linear dimensions of th
QveST [7]

we know that

the volume of a cone is equal to

V= \frac{1}{3} \pi r^{2}h

in this problem

the radius is equal to

r= \frac{112}{2}= 56ft

1) <u>Find the height of the cone before the storm</u>

Applying the Pythagorean Theorem find the height

h^{2} = l^{2}-r^{2}

l=65 ft

h^{2} = 65^{2}-56^{2}

h^{2} = 1,089

h=33 ft

2) <u>Find the volume before the storm</u>

V= \frac{1}{3}*\pi* 56^{2}*33

V=34,496\pi\ ft^{3}

3) <u>Find the volume after the storm</u>

After a storm, the linear dimensions of the pile are 1/3 of the original dimensions

so

r=(56/3) ft

h=(33/3)=11 ft

V= \frac{1}{3}*\pi* (56/3)^{2}*11

V= 1,277.63\pi\ ft^{3}

<u>4) Find how this change affect the volume of the pile</u>

Divide the volume after the storm by the volume before the storm

\frac{1,277.63 \pi }{34,496 \pi } = \frac{1}{27}

therefore

<u>the answer part a) is</u>

The volume of the pile after the storm is \frac{1}{27} times the original volume

<u>Part b)</u>  Estimate the number of lane miles that were covered with salt

5) <u>Find the amount of salt that was used during the storm</u>

=34,496 \pi - 1,277.63 \pi \\= 33.218.37 \pi \\= 104,358.59\ ft^{3}

6) <u>Find the pounds of road salt used</u>

104,358.59*80=8,348,687.2\ pounds    

7) <u>Find the number of lane miles that were covered with salt</u>

8,348,687.2/350=23,853.39 \ lane\ miles  

therefore

<u>the answer part b) is</u>

the number of lane miles that were covered with salt is 23,853.39 \ lane\ miles

<u>Part c) </u>How many lane miles can be covered with the remaining salt? Round your answer to the nearest lane mile

the remaining salt is equal to 1,277.63\pi\ ft^{3}

1,277.63\pi\ ft^{3}=4,013.79\ ft^{3}

8) <u>Find the pounds of road salt </u>

4,013.79*80=321,103.20\ pounds

9) <u>Find the number of lane miles </u>

321,103.20/350=917.44 \ lane\ miles

therefore

<u>the answer part c) is</u>

the number of lane miles is 917 \ lane\ miles

7 0
2 years ago
a school district requires all graduating seniors to take a mathematics test. This year, the rest scores were approximately norm
Klio2033 [76]
A score of 85 would be 1 standard deviation from the mean, 74.  Using the 68-95-99.7 rule, we know that 68% of normally distributed data falls within 1 standard deviation of the mean.  This means that 100%-68% = 32% of the data is either higher or lower.  32/2 = 16% of the data will be higher than 1 standard deviation from the mean and 16% of the data will be lower than 1 standard deviation from the mean.  This means that 16% of the graduating seniors should have a score above 85%.
3 0
2 years ago
Triangle L N M is shown. Angle L N M is a right angle. An altitude is drawn from point N to point O on side L M, forming a right
zaharov [31]

Answer:

Therefore the value of k = 6.

Step-by-step explanation:

Given:

LN = m

NM = l

OM = k

NO = 4

LO = 8

LM = 8 + k and

Δ LNM ,Δ LON and Δ MON are right Triangle.

To Find :

Om = k = ?

Solution:

In Right angle Triangle By Pythagoras Theorem we have,

(\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}

So, In Right angle Triangle Δ LON we have,

LN² = ON² + OL²

m² = 4² + 8²

m² = 80   ............( 1 )

Now in  Right angle Triangle Δ MON we have,

MN² =  ON² + MO²

l² = 4² + k² ....................( 2 )

Now In Right angle Triangle Δ LNM we have,

LM² = LN² + MN²

(8 + k)² = m² + l² .................( 3 )

Substituting equation  1 and equation 2 in equation 3

(8+k)² = 80 + 4² + k²

Applying (A+B)² = A² +2AB + B² we get

64 + 16k+k^{2} = 80+ 16 +k^{2} \\\\16k=96\\\\\therefore k=\frac{96}{16} \\\\\therefore k=6

Therefore the value of k = 6.

4 0
1 year ago
Read 2 more answers
Jose rides his bicycle for 5 minutes to travel 8 blocks. He rides for 10 minutes to travel 16 blocks.
kolbaska11 [484]

Answer:

A=8 B=15 C=40

Step-by-step explanation:

I just took a test

4 0
1 year ago
Read 2 more answers
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