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3 years ago

Ex 1) A population of 422, 000 increases by 12% each year. What is the population after years.

Mathematics

1 answer:

SpyIntel [72]3 years ago

4 0

Answer:472640

Step-by-step explanation:

Increase=112/100×422000

=472640

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5. If y= x+ 5 is an equation representing a line on a graph,<br> what is the slope (m)?<br> m<br> =

aliina [53]

Answer:

m = x

Step-by-step explanation:

y = mx + b

Mx will always be slope if the equation.

7 0

3 years ago

Given that f(x)=x^2-7x-1, g(x)=2x-3, and h(x)=4x-5 find each function. (f+g)(x)

Alexxandr [17]

Answer:

x^2-7x-1+2x-3

x^2-5x-4

5 0

3 years ago

How to solve please help thank you

Rashid [163]

her new balance is $105.

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7 0

3 years ago

Read 2 more answers

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

MrMuchimi

Answer:

$\large\boxed{1.\ f^{-1}(x)=4\log(x\sqrt[4]2)}\\\\\boxed{2.\ f^{-1}(x)=\log(x^5+5)}\\\\\boxed{3.\ f^{-1}(x)=\sqrt{4^{x-1}}}$

Step-by-step explanation:

$\log_ab=c\iff a^c=b\\\\n\log_ab=\log_ab^n\\\\a^{\log_ab}=b\\\\\log_aa^n=n\\\\\log_{10}a=\log a\\=============================$

$1.\\y=\left(\dfrac{5^x}{2}\right)^\frac{1}{4}\\\\\text{Exchange x and y. Solve for y:}\\\\\left(\dfrac{5^y}{2}\right)^\frac{1}{4}=x\qquad\text{use}\ \left(\dfrac{a}{b}\right)^n=\dfrac{a^n}{b^n}\\\\\dfrac{(5^y)^\frac{1}{4}}{2^\frac{1}{4}}=x\qquad\text{multiply both sides by }\ 2^\frac{1}{4}\\\\\left(5^y\right)^\frac{1}{4}=2^\frac{1}{4}x\qquad\text{use}\ (a^n)^m=a^{nm}\\\\5^{\frac{1}{4}y}=2^\frac{1}{4}x\qquad\log_5\ \text{of both sides}$

$\log_55^{\frac{1}{4}y}=\log_5\left(2^\frac{1}{4}x\right)\qquad\text{use}\ a^\frac{1}{n}=\sqrt[n]{a}\\\\\dfrac{1}{4}y=\log(x\sqrt[4]2)\qquad\text{multiply both sides by 4}\\\\y=4\log(x\sqrt[4]2)$

$--------------------------\\2.\\y=(10^x-5)^\frac{1}{5}\\\\\text{Exchange x and y. Solve for y:}\\\\(10^y-5)^\frac{1}{5}=x\qquad\text{5 power of both sides}\\\\\bigg[(10^y-5)^\frac{1}{5}\bigg]^5=x^5\qquad\text{use}\ (a^n)^m=a^{nm}\\\\(10^y-5)^{\frac{1}{5}\cdot5}=x^5\\\\10^y-5=x^5\qquad\text{add 5 to both sides}\\\\10^y=x^5+5\qquad\log\ \text{of both sides}\\\\\log10^y=\log(x^5+5)\Rightarrow y=\log(x^5+5)$

$--------------------------\\3.\\y=\log_4(4x^2)\\\\\text{Exchange x and y. Solve for y:}\\\\\log_4(4y^2)=x\Rightarrow4^{\log_4(4y^2)}=4^x\\\\4y^2=4^x\qquad\text{divide both sides by 4}\\\\y^2=\dfrac{4^x}{4}\qquad\text{use}\ \dfrac{a^n}{a^m}=a^{n-m}\\\\y^2=4^{x-1}\Rightarrow y=\sqrt{4^{x-1}}$

6 0

4 years ago

Two alloys A and B are being used to manufacture a certain steel product. An experiment needs to be designed to compare the two

sveta [45]

Answer:

Step-by-step explanation:

From the given information, we can compute the table showing the summarized statistics of the two alloys A & B:

Alloy A Alloy B

Sample mean $\bar {x} _A = 49.5$ $\bar {x} _B = 45.5$

Equal standard deviation $\sigma_A = 5$ $\sigma_B= 5$

Sample size $n_A = 30$ $n_{B}= 30$

Mean of the sampling distribution is :

$\mu_{\bar{X_1}-\bar{X_2}}= 49.5-45.5 \\ \\ = 4.0$

Standard deviation of sampling distribution:

$\sigma_{\bar{x_1}-\bar{x_2} }= \sqrt{\sigma^2_{\bar{x_1}-\bar{x_2} }} \\ \\ = \sqrt{\dfrac{\sigma_1^2}{n_1} + \dfrac{\sigma^2_2}{n_2} } \\ \\ =\sqrt{\dfrac{25}{30}+\dfrac{25}{30}} \\\\=\sqrt{1.667} \\ \\ =1.2909$

Hypothesis testing.

Null hypothesis: $H_o: \mu_A -\mu_B = 0$

Alternative hypothesis: $H_A:\mu_A -\mu_B > 4$

The required probability is:

$P(\overline X_A - \overline X_B>4|\mu_A - \mu_B) = P\Big (\dfrac{(\overline X_A - \overline X_B)-\mu_{X_A-X_B}}{\sigma_{\overline x_A -\overline x_B}} > \dfrac{4 - \mu_{X_A-\overline X_B}}{\sigma _{\overline x_A - \overline X_B}} \Big) \\ \\ = P \Big( z > \dfrac{4-0}{1.2909}\Big) \\ \\ = P(z \ge 3.10)\\ \\ = 1 - P(z < 3.10) \\ \\ \text{Using EXCEL Function:} \\ \\ = 1 - [NORMDIST(3.10)] \\ \\ = 1- 0.999032 \\ \\ 0.000968 \\ \\ \simeq 0.0010$

This implies that a minimal chance of probability shows that the difference of 4 is not likely, provided that the two population means are the same.

Since the P-value is very small which is lower than any level of significance.

Then, we reject $H_o$ and conclude that there is enough evidence to fully support alloy A.

3 0

3 years ago