Write an equation for a line perpendicular to the x-axis that passes through the point (5,1)

The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a

nadya68 [22]

Answer:

Initial amount is 523 grams

Amount after 100 years is 52 grams

Step-by-step explanation:

General exponential function for radioactive substances is given by

A(t) = Ao(1/2)^t/t1/2

Where,

A(t) is the amount remaining after t years

Ao is the initial amount of the radioactive substance

t is the number of years required for the radioactive substance to decay to A

t1/2 is the half-life of the radioactive substance

Comparing A(t) = 523(1/2)^t/30 with the general exponential function

Initial amount (Ao) = 523 grams

A(t) = 523(1/2)^t/30

t = 100

A(100) = 523(1/2)^100/30 = 523×0.5^3.333 = 523×0.0992 = 51.8816 = 52 grams (to the nearest gram)

7 0

3 years ago

9. A dust mite is about 250 microns long. One micron equals 0.001 millimeter. How long is a

lora16 [44]

Answer:

0.250

Step-by-step explanation:

0.001x250=0.250

6 0

3 years ago

Is the square root of 19 rational

kicyunya [14]

Answer:

No

Step-by-step explanation:

The square root needs to be a perfect square root to be a rational number, for example: the square root of 25 which equals 5.

3 0

3 years ago

In a study, researchers wanted to measure the effect of alcohol on the hippocampal region, the portion of the brain responsibl

schepotkina [342]

Answer:

$t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28$

The degrees of freedom are given by

$df=n-1=17-1=16$

The p value would be given by this probability:

$p_v =P(t_{(16)}$

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

Step-by-step explanation:

Information given

$\bar X=8.19 cm^3$ represent the sample mean

$s=0.8 cm^3$ represent the sample deviation

$n=17$ sample size

$\mu_o =9.02$ represent the value to verify

$\alpha=0.01$ represent the significance level for the hypothesis test.

t would represent the statistic

$p_v$ represent the p value

System of hypothesis to check

We want to check if the true mean is less than the normal value of 9.02 cm^3, the system of hypothesis would be:

Null hypothesis: $\mu \geq 9.02$

Alternative hypothesis: $\mu < 9.02$

The statistic for this case is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing the info given we got:

$t=\frac{8.19-9.02}{\frac{0.8}{\sqrt{17}}}=-4.28$

The degrees of freedom are given by

$df=n-1=17-1=16$

The p value would be given by this probability:

$p_v =P(t_{(16)}$

Since the p value is lower than the significance level provided of 0.01 we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly lower than 9.02 cm^3

6 0

3 years ago

How are natural numbers, whole numbers and integers used in everyday life?

Nezavi [6.7K]

Natural numbers: Counting things! You look around your room and see an electronic device, then another, then another! You just counted to 3 using the natural numbers.

Whole numbers! You try to look for electronic devices and realise that they’re all gone. You have zero electronic devices, and you just used whole numbers.

You go online to find where your electronic devices went, and realise they were taken because you’re in debt to the bank so they took some of your stuff. You’re in negative numbers, and now you’ve used integers.

4 0

3 years ago