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3 years ago

What is the volume of this container?HELPPPPP

Mathematics

1 answer:

worty [1.4K]3 years ago

4 0

Answer:

(2×4×4)+(3×2×4)

56 is the answer

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4. Bobby is 12 years old. His grandfather is 60 years old, what percent of Bobby's

Lina20 [59]

Answer: 12/60 as a percentage is 20%

Step-by-step explanation: boby is 12 and his grandfather is 60 so Bobby’s age is 12/60 of his grandfather and as a percentage it’s 20%

5 0

3 years ago

Read 2 more answers

Write an equation of the line that passes through (−5, −2)(−5, −2) and is parallel to y=23x+1y=23x+1.

Klio2033 [76]

Assuming you meant 2/3
y=2/3x+1 and (-5,-2)

paralell means same slope
y=mx+b
m=slope

y=2/3x+1
sllope=2/3

point slope
y-y1=m(x-x1)
for slope=m
a point is (x1,y1)

slope=2/3
point=(-5,-2)

y-(-2)=2/3(x-(-5))
y+2=2/3(x+5)
or
y=2/3x+16/3

7 0

3 years ago

You want to know the number of students in your school who read some of the newspaper at least once a week you surveyed 30 rando

LiRa [457]

Answer:

Can you be more specific

Step-by-step explanation:

5 0

3 years ago

On a given planet, the weight of an object varies directly with the mass of the object. Suppose that an object whose mass is 8 k

gladu [14]

Set it up as a fraction:
8kg = 10kg
24N ?N
cross multiply
8x=240
x=30N

8 0

3 years ago

Consider a Poisson distribution with μ = 6.

bearhunter [10]

Answer:

a) $P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b) f(2) = 0.04462

c) f(1) = 0.01487

d) $P(X \geq 2) = 0.93803$

Step-by-step explanation:

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of successes

e = 2.71828 is the Euler number

$\mu$ is the mean in the given interval.

In this question:

$\mu = 6$

a. Write the appropriate Poisson probability function.

Considering $\mu = 6$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

b. Compute f (2).

This is P(X = 2). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

So f(2) = 0.04462

c. Compute f (1).

This is P(X = 1). So

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

So f(1) = 0.01487.

d. Compute P(x≥2)

This is:

$P(X \geq 2) = 1 - P(X < 2)$

In which:

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2)$

$P(X = x) = \frac{e^{-6}*6^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-6}*6^{0}}{(0)!} = 0.00248$

$P(X = 1) = \frac{e^{-6}*6^{1}}{(1)!} = 0.01487$

$P(X = 2) = \frac{e^{-6}*6^{2}}{(2)!} = 0.04462$

Then

$P(X < 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.00248 + 0.01487 + 0.04462 = 0.06197$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.06197 = 0.93803$

$P(X \geq 2) = 0.93803$

5 0

3 years ago