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3 years ago

F(x) = x^2 + 2x – 4 f(3) = ?

Mathematics

1 answer:

Ostrovityanka [42]3 years ago

8 0

Answer:

Step-by-step explanation:

f(3)=(3)^2+2(3)-4

f(3)=9+2(3)-4

f(3)=9+6-4

f(3)=15-4

f(3)=11

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Remember to show work and explain. Use the math font.

vichka [17]

Answer:

$\large\boxed{1.\ f^{-1}(x)=\sqrt[12]{3^x}}\\\\\boxed{2.\ f^{-1}(x)=\sqrt[4]{3^x}}\\\\\ \boxed{3.\ f^{-1}(x)=\sqrt[3]{4^{7-x}}}$

Step-by-step explanation:

$(a^n)^m=a^{nm}\\\\\log_ab=c\iff a^c=b\\\\a^{\log_ax}=x\\\\n\log_ab=\log_ab^n\\\\\log_ab+\log_ac=\log_a(bc)\\============================\\\\1.\\y=3\log_3x^4\to y=\log_3(x^4)^3\to y=\log_3x^{12}$

$2.\\y=\log_3x^4\\\\\text{Exchange x and y. Solve for y:}\\\\\log_3y^4=x\Rightarrow3^{\log_3y^4}=3^x\Rightarrow y^{4}=3^x\\\\y=\sqrt[4]{3^x}\\-------------------------$

$3.\\y=-\log_4x^3+7\\\\\text{Exchange x and y. Solve for y:}\\\\-\log_4y^3+7=x\qquad\text{subtract 7 from both sides}\\\\-\log_4 y^3=x-7\qquad\text{change the signs}\\\\\log_4y^3=7-x\Rightarrow4^{\log_4y^3}=4^{7-x}\\\\y^3=4^{7-x}\Rightarrow y=\sqrt[3]{4^{7-x}}$

7 0

2 years ago

If 3 3/4 lb of candy costs 20 .25 how many would 1 lb of candy cost

Mrac [35]

solution:

$\frac{c}{w}=\frac{20.25}{\frac{3}{4}}\\\frac{c}{1}=\frac{20.25\times 4}{3}\\\frac{c}{1}=\frac{81}{3}\\c=27\\1Ib cost is \$ 27$

3 0

3 years ago

6, 9, 12, 15 which pair of numbers has an odd product and an even sum?

amid [387]

The only way for two integers to have an odd product is each integer is odd.
For example, 1*3=3 (odd), 1 and 3 are both odd.
Or,
5*11=55, all of 5,11,55 are odd.
The sum of two odd integers is always even, so the condition of even sum is automatically satisfied when the product is odd.
Out of the four integers, there are only two odd numbers, so choose the pair to be these two odd numbers and you'd get the right answer.

8 0

2 years ago

Suppose that a movie theater snack bar turns over its inventory of candy 3.3 times per month. (Round your answer to 2 decimal pl

hammer [34]

Answer:

The average daily sales rate for candy is 37.4.

Step-by-step explanation:

We know that,

$\text{Inventory turnover }= \frac{\text{Total sale}}{\text{Average inventories}}$

$\implies \text{Total sale}=\text{Inventory turnover }\times \text{Average inventories}$

Given,

Inventory turnover of candies = 3.3,

Average inventories = 340

So, sale of candies = 3.3 × 340 = 1122

Now,

$\text{Average daily sales rate}=\frac{\text{total sale}}{\text{Number of days}}$

Since, 1 month = 30 days ( approx ),

Hence, the average daily sales rate for candy = $\frac{1122}{30}$ =37.4

6 0

3 years ago

There are two major cell phone providers in the Colorado Springs, Colorado area, one called HTC and the other, Mountain Communic

Sergeeva-Olga [200]

Answer:

Step-by-step explanation:

Hello!

The objective of this exercise is to compare the "Churn rate" between the two major cell phone providers in colorado springs.

For each provider, the number of subscribers was noted at the beginning and end of a month.

The study variables are:

X₁: Number of customers that cut ties with HTC over a month.

X₂: Number of customers that cut ties with Mountain Communications over a month.

To compare the churn rate of both companies, the parameters of interest are the population proportions of customers that stopped using the companies services over the given period.

The hypotheses of interest are:

H₀: p₁ = p₂

H₁: p₁ ≠ p₂

α: 0.01

The statistic to use is the standard normal approximation:

$Z= \frac{('p_1-'p_2)- (p_1-p_2)}{\sqrt{'p(1 - 'p)[\frac{1}{n_1} + \frac{1}{n_2} ]} } }$

Z≈N(0;1)

HTC

n₁= 10000 customers (beginning of the month)

x₁= 190 customers (people that stopped using the company service)

sample proportion 'p₁= x₁/n₁= 190/10000= 0.019

Mountain Communications

n₂= 12285 customers (beginning of the month)

x₂= 215 customers (people that stopped using the company service)

sample proportion 'p₂= x₂/n₂= 215/12285= 0.0175

pooled sample proportion $'p= \frac{x_1 + x_2}{n_1 + n_2} = \frac{190 + 215}{10000 + 12285}$

'p= 0.0182

$Z_{H_0}= \frac{(0.019 - 0.0175) - 0}{\sqrt{(0.0182*0.9818)[\frac{1}{10000} + \frac{1}{12285} ]} }$

$Z_{H_0}= 0.833$

The p-value for this test is: 0 .4048

Note: Using the p-value method to decide over a hypothesis test, the decision rule is:

If p-value ≥ α, then you do not reject the null hypothesis.

If p-value < α, then you reject the null hypothesis.

The p-value < α, so the decision is to reject the null hypothesis.

With a significance level of 1%, there is enough evidence to conclude that there is a difference between the churn rate of the two cell phone providers.

I hope it helps!

3 0

2 years ago