List the factors of 15. list the multiples through 15 of each number

Simplify each exponential expression using the properties of exponents and match it to the correct answer.

saveliy_v [14]

1) (2 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2 = 2

2) (3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2) = 1/2

3) (3^7 $\times$ 4^7) (2 $\times$ 5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2

4) (2.3)^-1 (2^0) / (2.3)^-1 = 1

<u>Step-by-step explanation</u>:

Step 1 :

(2 $\times$ 3^-2)^3 (5 $\times$ 3^2)^2 / (3^-2)(5 $\times$ 2)^2

⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)

⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)

⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)

⇒ 2

Step 2 :

(3^3) (4^0)^2 (3 $\times$ 2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

⇒ 2^-1 = 1/2

Step 3 :

(3^7 $\times$ 4^7) (2 $\times$ 5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)

⇒ (2^-3) (5^-1) / (5^-1) (2^-4)

⇒ 1 / (2^-1)

⇒ 2

Step 4 :

(2.3)^-1 (2^0) / (2.3)^-1

⇒ (2^0)

⇒ Any number with power 'zero' is 1

⇒ 1

8 0

3 years ago

Which of the following modifications to the list of assets and liabilities below would result in an increase in net worth?

jolli1 [7]

The modifications made to the list of assets and liabilities that would result in an increase in net worth is Option D. Net worth is defined as the value of all the non-financial and financial assets owned by an individual or an institution.

All of the above (Putting $100 in savings, Paying $100 on credit cards and Getting paid $100)

5 0

3 years ago

Read 2 more answers

W + 15 = 29 <br> w=___ ?? HELP

kykrilka [37]

Answer:

w=14

Step-by-step explanation:

29-15=14

7 0

3 years ago

Read 2 more answers

An automobile manufacturer has given its car a 46.7 miles/gallon (MPG) rating. An independent testing firm has been contracted t

erastova [34]

Answer:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

Step-by-step explanation:

Information given

$\bar X=46.5$ represent the mean

$\sigma=1.1$ represent the population standard deviation

$n=150$ sample size

$\mu_o =46.7$ represent the value to verify

$\alpha=0.05$ represent the significance level for the hypothesis test.

z would represent the statistic

$p_v$ represent the p value

Hypothesis to test

We want to test if the true mean for this case is 46.7, the system of hypothesis would be:

Null hypothesis: $\mu = 46.7$

Alternative hypothesis: $\mu \neq 46.7$

Since we know the population deviation the statistic is given by:

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)

Replacing we got:

$z=\frac{46.5-46.7}{\frac{1.1}{\sqrt{150}}}=-2.23$

The p value would be given by:

$p_v =2*P(z$

For this case since th p value is lower than the significance level of0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true mean for this case is significantly different from 46.7 MPG

7 0

3 years ago

A carpenter is constructing staircase in a house. The distance from the floor to the basement is 10.8 feet. The staircase will b

konstantin123 [22]

Answer:

$\theta = 40.59^\circ$