PLEASE HELP! DUE TONIGHT

Softa [21]

Simplifying
j = (2j + 3)

Reorder the terms:
j = (3 + 2j)

Remove parenthesis around (3 + 2j)
j = 3 + 2j

Solving
j = 3 + 2j

Solving for variable 'j'.

Move all terms containing j to the left, all other terms to the right.

Add '-2j' to each side of the equation.
j + -2j = 3 + 2j + -2j

Combine like terms: j + -2j = -1j
-1j = 3 + 2j + -2j

Combine like terms: 2j + -2j = 0
-1j = 3 + 0
-1j = 3

Divide each side by '-1'.
j = -3

Simplifying
j = -3

8 0

3 years ago

Find the fifth term in the sequence that is defined as follows:

hram777 [196]

Answer : The fifth term in the sequence is -2.

Step-by-step explanation :

As we are given that the expression to calculate the $n^{th}$ term.

The expression is as follows:

$a_n=2(-1)^n$

where,

n is the number of term

Given:

n = 5

Now putting the value of n in the above expression, we get:

$a_n=2(-1)^n$

$a_5=2(-1)^5$

$a_5=2\times (-1)$

$a_5=-2$

Therefore, the fifth term in the sequence is -2.

5 0

3 years ago

Describe how the graph of y = x2 can be transformed to the graph of the given equation. y = (x+4)2

NNADVOKAT [17]

Moving 4 units to the left would change from y=x^2 to y=(x+4)^2.

6 0

3 years ago

Individuals filing federal income tax returns prior to March 31 had an average refund of $1102. Consider the population of "last

lions [1.4K]

Answer:

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) P-value = 0.0055

c) The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) Critical value tc=-1.96.

As t=-2.55, the null hypothesis is rejected.

Step-by-step explanation:

We have to perform a hypothesis test on the mean.

The claim is that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund ($1102).

a) The null hypothesis states that the last-minute filers average refund is equal to the early filers refund. The alternative hypothesis states that the last-minute filers average refund is less than the early filers refund.

$H_0: \mu=1102\\\\H_a:\mu < 1102$

b) The sample has a size n=600, with a sample refund of $1050 and a standard deviation of $500.

We can calculate the z-statistic as:

$t=\dfrac{\bar x-\mu}{s/\sqrt{n}}=\dfrac{1050-1102}{500/\sqrt{600}}=\dfrac{-52}{20.41}=-2.55$

The degrees of freedom are df=599

$df=n-1=600-1=599$

The P-value for this test statistic is:

$P-value=P(t$

c) Using a significance level α=0.05, the P-value is lower than the significance level, so the effect is significant. The null hypothesis is rejected.

There is enough statistical evidence to support the claim that that the refunds of the individuals that wait until the last five days to file their returns is on average lower than the early filers refund.

d) If the significance level is α=0.025, the critical value for the test statistic is t=-1.96. If the test statistic is below t=-1.96, then the null hypothesis should be rejected.

This is the case, as the test statistic is t=-2.55 and falls in the rejection region.

4 0

4 years ago

Really need help with this question

saw5 [17]

Keep the bases of 6 and add exponents: $6^{16}$

Keep the bases of 11 and add exponents: $11^{26}$

So its C.

hope that helps :)

3 0

3 years ago