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Snezhnost [94]
2 years ago
14

If you were babysitting, which would you rather: Charge $5 for the first hour and $8 for each additional hour? Charge $15 for th

e first hour and $6 for each additional hour? Explain your reasoning. Students, draw anywhere on this slide
(show your work plz )​
Mathematics
1 answer:
daser333 [38]2 years ago
5 0

Answer:

cahrge 15 for first hours and 6 for every additional hour

Step-by-step explanation:

you get more money because if you do one hour then you will still get $15 in a hour while if you do one hour in the other equation yolu will only get $5

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5x7/8 less than or greater
Pavel [41]
5x7= 35
35> (greater than) 8
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3 years ago
Read 2 more answers
Solve for h v=pie*r^2h
Elanso [62]

Answer:

v=πr^2h

v/πr^2=h

Step-by-step explanation:

4 0
3 years ago
What percent of 72 is 9? 1/8% 1 1/4% 12 1/2% 125%
USPshnik [31]

Answer:

12\dfrac 12\%

Step by step explanation:

\text{Let it be}~ x\%,\\\\~~~~~72 \cdot x \% = 9\\\\\\\implies 72\cdot \dfrac{x}{100} = 9\\\\\\\implies x = \dfrac{9 \times 100}{72}\\\\\\\implies x = \dfrac{100}{8}\\\\\\\implies x = \dfrac{25}2\\\\\\\implies x = 12 \dfrac 12\\\\\\\text{Hence 9 is}~ 12\dfrac 12\% ~ \text{of}~ 72

4 0
2 years ago
Assume that when adults with smartphones are randomly selected, 54% use them in meetings or classes (based on data from an lg sm
GuDViN [60]
Answer: 0.951%

Explanation:

Note that in the problem, the scenario is either the adult is using or not using smartphones. So, we have a yes or no scenario involved with the random variable, which is the number of adults using smartphones. Thus, the number of adults using smartphones follows the binomial distribution.

Let x be the number of adults using smartphones and n be the number of randomly selected adults. In Binomial distribution, the probability that there are k adults using smartphones is given by

P(x = k) = \frac{n!}{k!(n-k)!}p^k (1-p)^{n-k}

Where p = probability that an adult is using smartphones = 54% (since 54% of adults are using smartphones). 

Since n = 12 and k = 3, the probability that fewer than 3 are using smartphones is given by

P(x \ \textless \  3) = P(x = 0) + P(x = 1) + P(x = 2)
\\ \indent = \frac{12!}{0!(12-0)!}(0.54)^0 (1-0.54)^{12-0} + \frac{12!}{1!(12-1)!}(0.54)^1 (1-0.54)^{12-1} + \\ \indent \frac{12!}{2!(12-2)!}(0.54)^2 (1-0.54)^{12-2}
\\
\\ \indent = \frac{12!}{(1)(12!)}(0.46)^{12} + \frac{12(11!)}{(1)(11!)}(0.54)(0.46)^{11}+ \frac{12(11)(10!)}{(2)(10!)}(0.54)^2(0.46)^{10}
\\
\\ \indent = (1)(0.46)^{12} + (12)(0.54)(0.46)^{11}+ (66)(0.54)^2(0.46)^{10}
\\ \indent \boxed{P(x \ \textless \  3) \approx 0.00951836732 }


Therefore, the probability that there are fewer than 3 adults are using smartphone is 0.00951 or 0.951%.


5 0
3 years ago
What is the measure of z?<br> an<br> z= [?
Gemiola [76]

what to find

measure of z =1

7 0
3 years ago
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