Ask question

Ask question

All categories

2 years ago

6

Simplify the expression (2^3x5^3) x ( 2x5^4)^3 Fill in the blank with the correct values. PLZ if u know how to do this PLZ give

me the right answers.
(2^3x5^3)^2x(2x5^4)^3=2 Blank Blank
Blank

1 answer:

Dmitry [639]2 years ago

4 0

Answer:

244140625000

Step-by-step explanation:

You might be interested in

Least common factor of 9/24

adoni [48]

The answer to this is: 8

7 0

2 years ago

Escribe el decimal que sea 1/10 de 3.0

Misha Larkins [42]

¡creo que la respuesta sería .3!

6 0

3 years ago

The math club needs to raise more than $552. $.50 for a trip to state competition. The club has raise 12% of the funds which is

ratelena [41]

Each of the members must raise at least $69.46.

To start they need to raise $552.50. However, they have already raised 12% of it. They only need to raise 88% of it.

552.50 x 0.88 = 486.2

Divide the remaining amount by 7 to determine how much each individual must raise.

486.2 / 7 = 69.46

4 0

3 years ago

In a previous section we looked at mixing problems in which the volume of fluid remained constant and saw that such problems giv

Juli2301 [7.4K]

Answer:

Detailed solution is given below:

3 0

3 years ago

In analyzing the city and highway fuel efficiencies of many cars and trucks, the mean difference in fuel efficiencies for the 60

Olegator [25]

Answer:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

Step-by-step explanation:

Data provided

$\bar X=7.38$ represent the sample mean for the fuel efficiencies

$\mu$ population mean

s=2.51 represent the sample standard deviation

n=601 represent the sample size

Confidence interval

The formula for the confidence interval of the true mean is given by:

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

The degrees of freedom for this case is given by:

$df=n-1=601-1=600$

The Confidence level for this case is 0.95 or 95%, and the significance level $\alpha=0.05$ and $\alpha/2 =0.025$ and the critical value is given by $t_{\alpha/2}=1.96$

Replcing into the formula for the confidence interval is given by:

$7.38-1.96\frac{2.51}{\sqrt{601}}=7.18$

$7.38+1.96\frac{2.51}{\sqrt{601}}=7.58$

For this case the confidence interval is given by :

$7.18 \leq \mu \leq 7.58$

8 0

3 years ago