Plz help me. I need help fast

Which ones are true !!!!?????

MAXImum [283]

Answer:

both If's and a b

Step-by-step explanation:

I don't know for sure

7 0

2 years ago

The population of a town after t years is represented by the function f(t)=7,248(0.983)tf(t)=7,248(0.983)t.

lidiya [134]

Answer:

The population of the town is 0.983 times the population of the town in the previous year.

B is correct.

Step-by-step explanation:

We are given a function which represent population of a town after t years.

$f(t)=7248(0.983)^t$

It is an exponential function. Exponential function wither decease or increase it depends on factor.

$y=a\cdot b^x$

b is factor which decides factor of exponential function decrease or increase.

If b >1 then function increase

If 0<b<1 then function decrease.

If we see our problem $f(t)=7248(0.983)^t$

Here, b=0.983<1

Function would be decrease by factor of 0.983

Thus, The population of the town is 0.983 times the population of the town in the previous year.

7 0

2 years ago

Read 2 more answers

The Fish and Game Department stocked a lake with fish in the following proportions: 30% catfish, 15% bass, 40% bluegill, and 15%

vagabundo [1.1K]

Answer:

1) $\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

2) $p_v =P(\chi^2_{3}>16.313)=0.000978$

And we got the same decision reject the null hypothesis at 5% of significance.

Step-by-step explanation:

Previous concepts

The Chi-Square test of independence is used "to determine if there is a significant relationship between two nominal (categorical) variables". And is defined with the following statistic:

$\chi^2 =\sum_{i=1}^n \frac{(O-E)^2}{E}$

Where O rpresent the observed values and E the expected values.

State the null and alternative hypothesis

Null hypothesis: The distribution is 30% catfish, 15% bass, 40% bluegill, and 15% pike

Alternative hypothesis: The distribution is NOT 30% catfish, 15% bass, 40% bluegill, and 15% pike

The observed values are given by the table given:

Catfish =112, BAss = 95, Bluegill=210, Pike=83

Calculate the expected values

In order to calculate the expected values we can use the following formula for each cell of the table

$E = \% Grand total$

$E_{Catfish}=500*0.3=150$

$E_{Bass}=500*0.15=75$

$E_{Bluegill}=500*0.4=200$

$E_{Pike}=500*0.15=75$

Part 1: Calculate the statistic

$\chi^2 =\frac{(112-150)^2}{150}+\frac{(95-75)^2}{75}+\frac{(210-200)^2}{200}+\frac{(83-75)^2}{75}=16.313$

$\chi^2 =16.313$

Calculate the critical value

First we need to calculate the degrees of freedom given by:

$df= (categories-1)=(4-1)= 3$

Since the confidence provided is 95% the significance would be $\alpha=1-0.95=0.05$ and we can find the critical value with the following excel code: "=CHISQ.INV(0.95,3)", and our critical value would be $\chi^2_{crit}=7.815$