How to figure it out

Solve by adding and subtracting rational expressions. (3x)/(5xy^(4))+(6x^(2))/(15x^(2)y) Remember to show your work!

slamgirl [31]

Answer:

Step-by-step explanation:

$\frac{3x}{5xy^{4}}+\frac{6x^{2}}{15x^{2}y}=\frac{3}{5y^{4}}+\frac{6}{15y}\\\\=\frac{3}{5y^{4}}+\frac{2}{5y}\\\\=\frac{3}{5y^{4}}+\frac{2*y^{3}}{5y*y^{3}}\\\\=\frac{3}{5y^{4}}+\frac{2y^{3}}{5y^{4}}\\\\=\frac{3+2y^{3}}{5y^{4}}$

4 0

3 years ago

Estimate the sum. I will give 5 points and brainiest.

Naddik [55]

Answer:

57.29333333

Step-by-step explanation:

add everything up

8 0

3 years ago

Read 2 more answers

Elizabeth spends $14.99 each month on song downloads. At this amount, how much will Elizabeth spend on song downloads in 2 years

krok68 [10]

14.99 x 12 = 179.88
179.88 x 2 = 359.76.

She would spend 359.76 in 2 years

6 0

3 years ago

Read 2 more answers

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e

professor190 [17]

Answer:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

$n=15$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =16$ represent the sample variance

$\sigma^2_0 =25$ represent the value that we want to verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0

3 years ago

A woman invested 4 times as much as 5% as she did at 6%. The total amount of interest she earns is 1 year from both accounts is

Mnenie [13.5K]

Answer:

at 5% interest, she invested $8,000

at 6% interest, she invested $2,000

Step-by-step explanation:

Let the invested amount at 5% interest = x

Let the invested amount at 6% interest = y

x = 4y - - - - (1) (A woman invested 4 times as much as 5% as she did at 6%)

5% of x = 0.05x

6% of y = 0.06y

0.05x + 0.06y = 520 - - - - (2) (The total amount of interest she earns is $520)

putting the value of x gotten in equation (1) into eqn (2):

0.05x + 0.06y = 520

where x = 4y

∴ 0.05(4y) + 0.06y = 520

0.2y + 0.06y = 520

0.26y = 520

y = 520 ÷ 0.26

y = $2000

from eqn (1)

x = 4y

since y = 2000

x = 4 × 2000

x = $8000

7 0

3 years ago