What is the answer to this

the surface area of a cuboid is 95cm² and its lateral surface area is 63cm². find the area of its base

wel

Answer:

The Area of the base is 16 cm² .

Step-by-step explanation:

Given as :

The surface area of the cuboid = x = 95 cm²

The lateral surface area of the cuboid = y = 63 cm²

Let The Area of the base = z cm²

Now, Let The length of cuboid = l cm

The breadth of cuboid = b cm

The height of cuboid = h cm

According to question

∵ The surface area of the cuboid = 2 ×(length × breadth + breadth × height + height × length)

Or, x = 2 ×(l × b + b × h + h × l)

Or, 95 = 2 ×(l × b + b × h + h × l)

Or, (l × b + b × h + h × l) = $\dfrac{95}{2}$ ....1

Similarly

∵lateral surface area of the cuboid = 2 ×(breadth × height + length × height)

Or, y = 2 ×(b × h + l × h)

Or, 2 ×(b × h + l × h) = 63

Or, (b × h + l × h) = $\dfrac{63}{2}$ ......2

Putting value of eq 2 into eq 1

so, (l × b + $\dfrac{63}{2}$ ) = $\dfrac{95}{2}$

Or, l × b = $\dfrac{95}{2}$ - $\dfrac{63}{2}$

Or, l × b = $\dfrac{95 - 63}{2}$

i.e l × b = $\dfrac{32}{2}$

so, l × b = 16

Now, Again

∵The Area of the base = ( length × breadth ) cm²

So, z = l × b

i.e z = 16 cm²

So, The Area of the base = z = 16 cm²

Hence,The Area of the base is 16 cm² . Answer

5 0

3 years ago

12=x+3. SOLVE FOR X NEED HELP ASAP

dimaraw [331]

12 - 3 = x , 9 = x . hope this helps

5 0

3 years ago

Read 2 more answers

After how many years are the salaries offered by company a and company b the same

12345 [234]

24000 + 3000x = 30000 + 2400x
3000x - 2400x = 30,000 - 24,000
600x = 6000
x = 6000/600
x = 10 years

6 0

2 years ago

Suppose we are interested in analyzing the weights of NFL players. We know that on average, NFL players weigh 247 pounds with a

tigry1 [53]

Answer:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

$p_v =P(z$

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

Step-by-step explanation:

For this case we have the following data given:

$\bar X =237$ represent the sample mean

$\sigma = 47$ represent the population deviation

$n =30$ represent the sample size selected

$\mu_0 = 247$ represent the value that we want to test.

The standard error for this case is given by:

$SE= \frac{\sigma}{\sqrt{n}} = \frac{47}{\sqrt{30}}= 8.58$

For the 90% confidence the value of the significance is given by $\alpha=1-0.9 = 0.1$ and $\alpha/2 = 0.05$ so we can find in the normal standard distribution a quantile that accumulates 0.05 of the area on each tail and we got:

$z_{\alpha/2}= 1.64$

And the margin of error would be:

$ME= 1.64 *\frac{\sigma}{\sqrt{n}} = 1.64*\frac{47}{\sqrt{30}}= 14.073$

The confidence interval for this case would be given by:

$\bar X - ME = 237- 14.073 = 222.927$

$\bar X + ME = 237+ 14.073 = 251.073$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (a)

And on this case we have that ME =5 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

Replacing into formula (b) we got:

$n=(\frac{1.640(47)}{5})^2 =237.65 \approx 238$

So the answer for this case would be n=238 rounded up to the nearest integer

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 247 pounds, the system of hypothesis would be:

Null hypothesis: $\mu \geq 247$

Alternative hypothesis: $\mu$

Since we know the population deviation, is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

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

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$z=\frac{237-247}{\frac{47}{\sqrt{30}}}=-1.165$

P-value

Since is a left tailed test the p value would be:

$p_v =P(z$

Conclusion

If we compare the p value and the significance level given $\alpha=0.1$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can't conclude that the true mean is lower than 247 at 10% of significance.

5 0

2 years ago

**The following scenario is for #12, #13, and #14 on this test.** A school chorus has 90 sixth grade students and 75 seventh gra

givi [52]

Answer:

The music director can make 15 groups of performers with 6 sixth grade students and 5 seventh grade students in each group.

Step-by-step explanation:

A school chorus has 90 sixth-grade students and 75 seventh-grade students.

Factor these two numbers:

Find GCF(90,75):

Now,

Therefore, the music director can make 15 groups of performers with 6 sixth grade students and 5 seventh grade students in each group.

6 0

2 years ago