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zhenek [66]
3 years ago
9

Billy has 1 gallon of paint. He is going to pour it into a paint tray that measures 10 inches wide, 14 inches long, and 4 cm dee

p.
Mathematics
1 answer:
guapka [62]3 years ago
5 0
You need to state the question, what you're trying to find. By using my context clues, I'm not sure what you're asking but
Length x Width x Height so
10 x 14 = 140
140 x 4 = 260
260 = area of the paint tray
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Math help please thanks
Pepsi [2]
For most of these questions, all you would have to do is add the angles together and or subtract the larger angle from the smaller one to find the other angle. So in the first case.

You would add the angles, 26 degrees and 60 degrees to find the angle KLM which is about 90 degrees a right angle, in this case it would be 86 degrees.

Similarly, for the other question, take the larger angle, 145 degrees and subtract 61 degrees to solve for the angle that will also add to give 145 degrees,in this case it would be 84 degrees.

3 0
3 years ago
How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?
lidiya [134]

Q. How many triangles can be constructed with sides measuring 5 m, 16 m, and 5 m?

Solution:

Here we are given with the sides of the triangle as 5m, 16m and 5.

As the Triangle inequality we know that

The sum of the length of the two sides should be greater than the length of the third side. But this inequality fails here.

Hence no triangle can be made.

So the correct option is None.

Q.How many triangles can be constructed with sides measuring 6 cm, 2 cm, and 7 cm?

Solution:

Here we are given with the sides of the triangle as 6m, 2m and 7m.

As the Triangle inequality we know that

The sum of the length of the two sides should be greater than the length of the third side. The given values follows the triangle inequality.

Hence one triangle can be formed.

So the correct option is  one.

8 0
3 years ago
!!!!!!!!!HELLLLLPPPP!!!!!!
myrzilka [38]

Answer:

maximum 4 Mimimum I'm 0 or maximum 10 minimum 0

4 0
2 years ago
Read 2 more answers
8
Nimfa-mama [501]
3
—
-2
because the answer is 6 over -4 and it simplified is that answer.
3 0
3 years ago
The port of South Louisiana, located along 54 miles of the Mississippi River between New Orleans and Baton Rouge, is the largest
Ksenya-84 [330]

Answer:

a) 0.7287

b) 0.9663

c) 0.237

d) 3.65 tons of cargo per week or more that will require the port to extend its operating hours.  

Step-by-step explanation:

We are given the following information in the question:

Mean, μ =  4.5 million tons of cargo per week

Standard Deviation, σ = 0 .82 million tons

We are given that the distribution of number of tons of cargo handled per week is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

a) P( port handles less than 5 million tons of cargo per week)

P(x < 5)

P( x < 5) = P( z < \displaystyle\frac{5 - 4.5}{0.82}) = P(z < 0.609)

Calculation the value from standard normal z table, we have,  

P(x < 5) =0.7287= 72.87\%

b) P( port handles 3 or more million tons of cargo per week)

P(x \geq 3) = P(z \geq \displaystyle\frac{3-4.5}{0.82}) = P(z \geq −1.82926)\\\\P( z \geq −1.82926) = 1 - P(z < -1.829)

Calculating the value from the standard normal table we have,

1 - 0.0337 = 0.9663 = 96.63\%\\P( x \geq 3) = 96.63\%

c)P( port handles between 3 million and 4 million tons of cargo per week)

P(3 \leq x \leq 4) = P(\displaystyle\frac{3 - 4.5}{0.82} \leq z \leq \displaystyle\frac{4-4.5}{0.82}) = P(-1.829 \leq z \leq -0.609)\\\\= P(z \leq -0.609) - P(z < -1.829)\\= 0.271-0.034 = 0.237= 23.7\%

P(3 \leq x \leq 4) = 23.7\%

d) P(X=x) = 0.85

We have to find the value of x such that the probability is 0.85.

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 4.5}{0.82})=0.85  

= 1 -P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.85  

=P( z \leq \displaystyle\frac{x - 4.5}{0.82})=0.15  

Calculation the value from standard normal z table, we have,  

P( z \leq -1.036) = 0.15

\displaystyle\frac{x - 4.5}{0.82} = -1.036\\x = 3.65

Thus, 3.65 tons of cargo per week or more that will require the port to extend its operating hours.

8 0
3 years ago
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