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stepladder [879]
3 years ago
5

A group of friends were working on a student film with a budget of $700, which they spent on costumes and equipment. They spent

$168 on equipment. What percent did they spend on costumes?
Mathematics
1 answer:
miskamm [114]3 years ago
5 0

Answer:

76%

Step-by-step explanation:

24% of 700=168

100-24=76

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Question 11<br>2, 10, 50, ...<br><br>plz help me out! ​
I am Lyosha [343]

Answer:

250

Step-by-step explanation:

Every number is multiplied by 5

2 x 5 = 10

10 x 5 = 50

50 x 5 = 250

7 0
3 years ago
A study of the US clinical population found that 24.5% are diagnosed with a mental disorder, 14% are diagnosed with an alcohol-r
Keith_Richards [23]

Answer:

Part (A) The probability that someone from the clinical population is diagnosed with a mental disorder, knowing that the person is diagnosed with an alcohol-related disorder is 0.286

Part (B) The probability that someone from the clinical population is diagnosed with an alcohol-related disorder, knowing that the person is diagnosed with a mental disorder is 0.163.

Step-by-step explanation:

Consider the provided information.

A study of the US clinical population found that 24.5% are diagnosed with a mental disorder, 14% are diagnosed with an alcohol-related disorder, and 4% are diagnosed with both disorders.

P(M) = P(Mental disorder) = 24.5% = 0.245  

P(A) =P(Alcohol-related disorder) = 14% = 0.14

P(M and A) =P(Both disorder) = 4% = 0.04

Part (A) What is the probability that someone from the clinical population is diagnosed with a mental disorder, knowing that the person is diagnosed with an alcohol-related disorder?

P(\text{Mental disorder}\mid\text{alcohol-related disorder})=\frac{P(M\ and\ A)}{P(A)}

P(\text{Mental disorder}\mid\text{alcohol-related disorder})=\frac{0.04}{0.14}=0.286

Hence, the required probability is 0.286.

Part (b) What is the probability that someone from the clinical population is diagnosed with an alcohol-related disorder, knowing that the person is diagnosed with a mental disorder.

P(\text{alcohol-related disorder}\mid\text{Mental disorder})=\frac{P(M\ and\ A)}{P(M)}

P(\text{alcohol-related disorder}\mid\text{Mental disorder})=\frac{0.04}{0.245}=0.163

Hence, the required probability is 0.163.

8 0
3 years ago
Please answer! I’ll give branliest! This is needed by tomorrow! Answers would be appreciated!
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Answer:

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Position 2 Top: Ne: Night, Winter. Nw: Day, Summer

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Step-by-step explanation:

8 0
3 years ago
Under certain conditions, the number of bacteria present in a colony is approximated by f(t) = Age 0.023t, where t is in minutes
AleksandrR [38]

Answer:

  • 5 min: 3,029,058
  • 10 min: 3,398,220
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Step-by-step explanation:

The given function is evaluated by substituting the given values of t. This requires using the exponential function of your calculator with a base of 'e'. Many calculators have that value built in, or have an e^x function (often associated with the Ln function).

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<h3>5 minutes</h3>

The number of bacteria present after 5 minutes is about ...

  f(5) = 2.7×10^6×e^(0.023×5) ≈ 3,029,058

<h3>10 minutes</h3>

The number of bacteria present after 10 minutes is about ...

  f(10) = 2.7×10^6×e^(0.023×10) ≈ 3,398,220

<h3>60 minutes</h3>

The number of bacteria present after 60 minutes is about ...

  f(60) = 2.7×10^6×e^(0.023×60) ≈ 10,732,234

8 0
2 years ago
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