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3 years ago

The table shows values for functions f(x) and g(x)

Mathematics

2 answers:

andriy [413]3 years ago

6 0

Answer:

Option 2nd and 3rd are correct.

The solutions to f(x) =g(x) are;

x = -2 and x = -1

Step-by-step explanation:

Let x represents the independent variable and f(x) and g(x) represents the dependent variable.

Given the function:

$f(x) = 2^(-x)$ and $g(x) = -2x$ \

we have to find the solution to f(x) = g(x).

From the given table you can see that:

For x = - 2

f(-2) = 4

g(-2) = 4

then;

f(x) = g(x) for x = -2

Also;

For x = - 1

f(-1) = 2

g(-1) = 2

then;

f(x) = g(x) for x =-1

Therefore, the solutions to f(x) =g(x) are, x = -1 and -2

Alecsey [184]3 years ago

5 0

Consider, pls, this solution:
f(x)=g(x) if x= -2 and x= -1, in those case f(x)=g(x)=4 and f(x)=g(x)=2

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Twice the sum of four and x is three less than x.

aliya0001 [1]

Answer:

-11

Step-by-step explanation

Condition:

2(4+x)=x-3

8+2x=x-3

Combining similar terms

8+3=x-2x

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7 0

2 years ago

A restaurant chain has two locations in a medium-sized town and, believing that it has oversaturated the market for its food, is

scoray [572]

Answer:

Yes, since the test statistic value is greater than the critical value.

Step-by-step explanation:

Data given and notation

$\bar X_{1}=360000$ represent the mean for the downtown restaurant

$\bar X_{2}=340000$ represent the mean for the freeway restaurant

$s_{1}=50000$ represent the sample standard deviation for downtown

$s_{2}=40000$ represent the sample standard deviation for the freeway

$n_{1}=36$ sample size for the downtown restaurant

$n_{2}=36$ sample size for the freeway restaurant

t would represent the statistic (variable of interest)

$\alpha=0.05$ significance level provided

Develop the null and alternative hypotheses for this study?

We need to conduct a hypothesis in order to check if the true mean of revenue for downtown is higher than for freeway restaurant, the system of hypothesis would be:

Null hypothesis: $\mu_{1}\leq \mu_{2}$

Alternative hypothesis: $\mu_{1} > \mu_{2}$

Since we don't know the population deviations for each group, for this case is better apply a t test to compare means, and the statistic is given by:

$t=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}$ (1)

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.

Determine the critical value.

Based on the significance level $\alpha=0.05$ and $\alpha/2=0.025$ we can find the critical values from the t distribution dith degrees of freedom df=36+36-2=55+88-2=70, we are looking for values that accumulates 0.025 of the area on the right tail on the t distribution.

For this case the value is $t_{1-\alpha/2}=1.66$

Calculate the value of the test statistic for this hypothesis testing.

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

$t=\frac{360000-340000}{\sqrt{\frac{50000^2}{36}+\frac{40000^2}{36}}}}=1.874$

What is the p-value for this hypothesis test?

Since is a right tailed test the p value would be:

$p_v =P(t_{70}>1.874)=0.033$

Based on the p-value, what is your conclusion?

Comparing the p value with the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we reject the null hypothesis, and the true mean for the downtown revenue restaurant seems higher than the true mean revenue for the freeway restaurant.