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ANTONII [103]
3 years ago
14

Tim worker is doing his budget. He discovers that the average miscellaneous expense is $45.00 with a standard deviation of $16.0

0
Mathematics
1 answer:
Marianna [84]3 years ago
3 0
So what's the question??

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What is the volume of a frustum of a rectangular pyramid if one base has measurements 6 ft. × 8 ft., the other base has measurem
RoseWind [281]
To solve this, we use the formula for the volume of the frustum of a cone which is expressed as:

<span>V= (H/6)[WL + (W+a)(L+b) + ab]
</span>V = 3/6 [(6 ft. × 8 ft.) + (6 + 14)(18 + 8) + (<span>14 ft. × 18 ft.)]
</span>V = 409.98 ft^3 <-----OPTION C 
7 0
3 years ago
Read 2 more answers
A clothing store offers a free T- shirt when a customer spends $75 or more. Lyndon has already spent $36.95 which statement best
Anna [14]
$38.05 or more because you need a minimum of $38.10
8 0
3 years ago
Which are rational and irrational? Thanks in advance :)
DochEvi [55]

Answer:

3\frac{1}{2}-rational

\pi -irrational

-\frac{2}{3}-rational

\sqrt{16} -rational

\sqrt{3} -irrational

18-rational

0.625-rational

0.987987987...-rational

Step-by-step explanation:

Rational numbers:

-are all numbers you can write as a quotient of integers  \frac{a}{b} , b\neq 0

-include terminating decimals. For example, \frac{1}{8}=0.125

-include repeating decimals. For example, \frac{1}{3} =0.33333....

Irrational numbers:

-have decimal representations that neither terminate nor repeat. For example, \sqrt{2}=1.414213....

-cannot be written as quotients of integers

8 0
3 years ago
Pls help me with this question I’m trying to Anwser
tino4ka555 [31]

Answer:Wish i could help, but I'm just as confused.

Step-by-step explanation:

8 0
4 years ago
Suppose the number of insect fragments in a chocolate bar follows a Poisson process with the expected number of fragments in a 2
leonid [27]

Answer:

a)The expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b)0.6004

c)19.607

Step-by-step explanation:

Let X denotes the number of fragments in 200 gm chocolate bar with expected number of fragments 10.2

X ~ Poisson(A) where \lambda = \frac{10.2}{200} = 0.051

a)We are supposed to find the expected number of insect fragments in 1/4 of a 200-gram chocolate bar

\frac{1}{4} \times 200 = 50

50 grams of bar contains expected fragments = \lambda x = 0.051 \times 50=2.55

So, the expected number of insect fragments in 1/4 of a 200-gram chocolate bar is 2.55

b) Now we are supposed to find the probability that you have to eat more than 10 grams of chocolate bar before ending your first fragment

Let X denotes the number of grams to be eaten before another fragment is detected.

P(X>10)= e^{-\lambda \times x}= e^{-0.051 \times 10}= e^{-0.51}=0.6004

c)The expected number of grams to be eaten before encountering the first fragments :

E(X)=\frac{1}{\lambda}=\frac{1}{0.051}=19.607 grams

7 0
3 years ago
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