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2 years ago

There are y boats on a lake.

Mathematics

2 answers:

alekssr [168]2 years ago

8 0

The answer is 7y! there are 7 people on each boat so you need to do 7 times a number that you don’t know which is y!

Doss [256]2 years ago

5 0

Answer:

The answer is 7y.

Thank you and please rate me as brainliest as it will help me to level up

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Thirty-three college freshmen were randomly selected for an on-campus survey at their university. The participants' mean GPA was

Ivan

Answer: $\pm0.1706$

Step-by-step explanation:

Given : Sample size : n= 33

Critical value for significance level of $\alpha:0.05$ : $z_{\alpha/2}= 1.96$

Sample mean : $\overline{x}=2.5$

Standard deviation : $\sigma= 0.5$

We assume that this is a normal distribution.

Margin of error : $E=\pm z_{\alpha/2}\dfrac{\sigma}{\sqrt{n}}$

i.e. $E=\pm (1.96)\dfrac{0.5}{\sqrt{33}}=\pm0.170596102837\approx\pm0.1706$

Hence, the margin of error is $\pm0.1706$

7 0

3 years ago

PLEAS HELP ASAP PLEAS 20 points<br> 1 over 3b = 4 over 5. what equals b in this equation?

White raven [17]

Answer:

b = 5/12

Step-by-step explanation:

You have the proportion

1/(3b) = 4/5

Cross multiply to get rid of the fractions...

5 = 12b now solve for b

5/12 = b (divide by 12 on both sides)

8 0

2 years ago

Suppose after 2500 years an initial amount of 1000 grams of a radioactive substance has decayed to 75 grams. What is the half-li

krok68 [10]

Answer:

The correct answer is:

Between 600 and 700 years (B)

Step-by-step explanation:

At a constant decay rate, the half-life of a radioactive substance is the time taken for the substance to decay to half of its original mass. The formula for radioactive exponential decay is given by:

$A(t) = A_0 e^{(kt)}\\where:\\A(t) = Amount\ left\ at\ time\ (t) = 75\ grams\\A_0 = initial\ amount = 1000\ grams\\k = decay\ constant\\t = time\ of\ decay = 2500\ years$

First, let us calculate the decay constant (k)

$75 = 1000 e^{(k2500)}\\dividing\ both\ sides\ by\ 1000\\0.075 = e^{(2500k)}\\taking\ natural\ logarithm\ of\ both\ sides\\In 0.075 = In (e^{2500k})\\In 0.075 = 2500k\\k = \frac{In0.075}{2500}\\ k = \frac{-2.5903}{2500} \\k = - 0.001036$

Next, let us calculate the half-life as follows:

$\frac{1}{2} A_0 = A_0 e^{(-0.001036t)}\\Dividing\ both\ sides\ by\ A_0\\ \frac{1}{2} = e^{-0.001036t}\\taking\ natural\ logarithm\ of\ both\ sides\\In(0.5) = In (e^{-0.001036t})\\-0.6931 = -0.001036t\\t = \frac{-0.6931}{-0.001036} \\t = 669.02 years\\\therefore t\frac{1}{2} \approx 669\ years$