Find the missing number so that the equation has infinitely many solutions.

Scientist can determine the age of ancient objects by a method called radiocarbon dating. The bombardment of the upper atmospher

Artemon [7]

Answer: 2500 years

Step-by-step explanation:

I'm not quite sure if I'm doing this right myself but I'll give it a shot.

We use this formula to find half-life but we can just plug in the numbers we know to find t.

$A(t)=A_{0}(1/2)^t^/^h$

We know half-life is 5730 years and that the parchment has retained 74% of its Carbon-14. For $A_{0$ let's just assume that there are 100 original atoms of Carbon-14 and for A(t) let's assume there are 74 Carbon-14 atoms AFTER the amount of time has passed. That way, 74% of the C-14 still remains as 74/100 is 74%. Not quite sure how to explain it but I hope you get it. h is the last variable we need to know and it's just the half-life, which has been given to us already, 5730 years, so now we have this.

$74=100(1/2)^t^/^5^7^3^0$

Now, solve. First, divide by 100.

$0.74=(0.5)^t^/^5^7^3^0$

Take the log of everything

$log(0.74)=\frac{t}{5730} log(0.5)$

Divide the entire equation by log (0.5) and multiply the entire equation by 5730 to isolate the t and get

$5730\frac{log(0.74)}{log(0.5)} =t$

Use your calculator to solve that giant mess for t and you'll get that t is roughly 2489.128182 years. Round that to the nearest hundred years, and you'll find the hopefully correct answer is 2500 years.

Really hope that all the equations that I wrote came out good and that that's right, this is definitely the longest answer I've ever written.

5 0

3 years ago

Based on the Nielsen ratings, the local CBS affiliate claims its 11:00 PM newscast reaches 41 % of the viewing audience in the a

ZanzabumX [31]

Answer:

1) Null hypothesis: $p\geq 0.41$

Alternative hypothesis: $p < 0.41$

2) $\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

3) $z_{crit}=-2.33$

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

4) $z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017$

5) $z_{crit}=-1.28$

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

6) We see that $|t_{calculated}|$ so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

7) Null hypothesis: $p\geq 0.41$

Step-by-step explanation:

Data given and notation

n=100 represent the random sample taken

X represent the people indicated that they watch the late evening news on this local CBS station

$\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

$p_o=0.41$ is the value that we want to test

$\alpha=0.05$ represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

$p_v{/tex} represent the p value (variable of interest) Part 1We need to conduct a hypothesis in order to test the claim that 11:00 PM newscast reaches 41 % of the viewing audience in the area: Null hypothesis:[tex]p\geq 0.41$

Alternative hypothesis: $p < 0.41$

Part 2

$\hat p=0.36$ estimated proportion of people indicated that they watch the late evening news on this local CBS station

Part 3

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.01 of the area on the left and on this case this value is :

$z_{crit}=-2.33$

And we can use the following excel code to find it: "=NORM.INV(0.01,0,1)"

Part 4

When we conduct a proportion test we need to use the z statisitc, and the is given by:

$z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}$ (1)

Since we have all the info requires we can replace in formula (1) like this:

$z=\frac{0.36 -0.41}{\sqrt{\frac{0.41(1-0.41)}{1000}}}=-1.017$

Part 5

Since we have a left tailed test we need to see in the normal standard distribution a value that accumulates 0.1 of the area on the left and on this case this value is :

$z_{crit}=-1.28$

And we can use the following excel code to find it: "=NORM.INV(0.1,0,1)"

Part 6

We see that $|t_{calculated}|$ so then we have enough evidence to FAIL to reject the null hypothesis at 1% of significance.

Part 7

Null hypothesis: $p\geq 0.41$

4 0

3 years ago

My friend has $729 to spend on a fence for her rectangular garden. She wants to use cedar fencing which costs $18/yard on one si

denis23 [38]

Answer:

two consecutive negative integers have a product of 110 . what are the integers?

Step-by-step explanation:

4 0

3 years ago

Initially, there were only 86 weeds in the garden. The weeds grew at a rate of 29% each week. The following function represents

Dmitry [639]

Answer:

Option A. $f(x)=86[1.04]^{x}$ ; grows approximately at a rate of 0.4% daily

Step-by-step explanation:

we have

$f(x)=86(1.29)^{x}$

where

f(x) the number of weeds in the garden

x ----> the number of weeks

Calculate how quickly the weeds grow each day

Remember that a week is equal to seven days

so

$f(x)=86(1.29)^{\frac{x}{7}}$

Using the law of exponents

b^(x/a) = b^(x*(1/a)) = (b^(1/a))^x

so

$f(x)=86[(1.29)^{\frac{1}{7}}]^{x}$

$f(x)=86[1.04]^{x}$

therefore

The rate is approximately

1.04=1+r

r=1.04-1=0.04=4% daily

8 0

3 years ago

Writeanequationforthelinethatpassesthrough(4,-3)and has a slope of 3.

djyliett [7]

Answer:

x+y=1 (if x=4 )

4+y=1

y=1-4

y=-3

7 0

3 years ago

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