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Tanzania [10]
3 years ago
8

Can you please help me​

Mathematics
1 answer:
MissTica3 years ago
4 0

Answer:

cream

Step-by-step explanation:

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Milton mixed 1/2 gallons of red paint , 2/5 gallons of blue paint, 1 2/3 gallons of orange paint into a mixing barrel. How many
never [62]
2 17/30 gallons of paint altogether

because 1/2 plus 2/5 plus 1 2/3 is 77/30 and if you simplify that it is 2 7/30
7 0
3 years ago
2/3x=10 help please and explain
stira [4]
 Solutions 

Simplify <span><span>2/3</span>x</span> to <span><span>2x/</span>3
</span>
<span><span><span>2x/</span>3</span>=10

</span>2) Multiply both sides by 3

<span>2x=10×3

3) </span>Simplify 10×3 to 30

<span>2x=30

4) </span>Divide both sides by <span>2

</span><span>x=30 </span>÷ <span>2

5) </span>Simplify <span>30 </span><span>÷ 2</span><span> </span><span>to</span><span> </span><span>15

</span><span><span>x=15

Answer = 15  

Also can be written as 1/15</span></span>
3 0
3 years ago
Read 2 more answers
Which of the following sets of numbers could not represent the three sides of a right
Kay [80]
The answer is B). {20,21,28}
4 0
3 years ago
Write three DIFFERENT properties that are equivalent to 3 4
stich3 [128]

Answer:

6/8

9/12

12/16

Step-by-step explanation:

6 0
3 years ago
Do you tailgate the car in front of you? About 35% of all drivers will tailgate before passing, thinking they can make the car i
Vesna [10]

Answer:

(a) The histogram is shown below.

(b) E (X) = 4.2

(c) SD (X) = 2.73

Step-by-step explanation:

Let <em>X</em> = <em>r</em><em> </em>= a driver will tailgate the car in front of him before passing.

The probability that a driver will tailgate the car in front of him before passing is, P (X) = <em>p</em> = 0.35.

The sample selected is of size <em>n</em> = 12.

The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 12 and <em>p</em> = 0.35.

The probability function of a binomial random variable is:

P(X=x)={n\choose x}p^{x}(1-p)^{n-x}

(a)

For <em>X</em> = 0 the probability is:

P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006

For <em>X</em> = 1 the probability is:

P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037

For <em>X</em> = 2 the probability is:

P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109

Similarly the remaining probabilities will be computed.

The probability distribution table is shown below.

The histogram is also shown below.

(b)

The expected value of a Binomial distribution is:

E(X)=np

The expected number of vehicles out of 12 that will tailgate is:

E(X)=np=12\times0.35=4.2

Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.

(c)

The standard deviation of a Binomial distribution is:

SD(X)=np(1-p)

The standard deviation of vehicles out of 12 that will tailgate is:

SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\

Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.

4 0
4 years ago
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