All categories

3 years ago

2

Mathematics

1 answer:

goldfiish [28.3K]3 years ago

4 0

Answer:

The volume of can A is half the volume of can B.

Step-by-step explanation:

Given

Can A and Can B

Required

The true statement

For Can A, we have:

$h =8$

$r = 0.5$

The volume is:

$Volume = \pi r^2h$

This gives:

$V_A = \pi * 0.5^2 * 8$

$V_A = \pi * 2$

$V_A = 2\pi$

For Can B, we have:

$h =4$

$r = 1$

The volume is:

$Volume = \pi r^2h$

This gives:

$V_B = \pi * 1^2 * 4$

$V_B = \pi * 4$

$V_B = 4\pi$

So, we have:

$V_A = 2\pi$

$V_B = 4\pi$

By comparison, (d) is correct

You might be interested in

at a local preschool there is a ratio of 4 of 3 boys to every 4 girls if there are 245 total preschoolers enrolled how many are

TiliK225 [7]

There are 105 boys. and 140 girls

4 0

3 years ago

A circular swimming pool has a diameter of 40 meters. The depth of the pool is constant along west-east lines and increases line

Nataliya [291]

Answer:

Volume is $2000\pi\ m^{3}$

Solution:

As per the question:

Diameter, d = 40 m

Radius, r = 20 m

Now,

From north to south, we consider this vertical distance as 'y' and height, h varies linearly as a function of y:

iff

h(y) = cy + d

Then

when y = 1 m

h(- 20) = 1 m

1 = c.(- 20) + d = - 20c + d (1)

when y = 9 m

h(20) = 9 m

9 = c.20 + d = 20c + d (2)

Adding eqn (1) and (2)

d = 5 m

Using d = 5 in eqn (2), we get:

$c = \frac{1}{5}$

Therefore,

$h(y) = \frac{1}{5}y + 5$

Now, the Volume of the pool is given by:

$V = \int h(y)dA$

where

A = $r\theta$

$A = rdr\ d\theta$

Thus

$V = \int (\frac{1}{5}y + 5)dA$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}rsin\theta + 5) rdr\ d\theta$

$V = \int_{0}^{2\pi}\int_{0}^{20} (\frac{1}{5}r^{2}sin\theta + 5r}) dr\ d\theta$

$V = \int_{0}^{2\pi} (\frac{1}{15}20^{3}sin\theta + 1000) d\theta$

$V = [- 533.33cos\theta + 1000\theta]_{0}^{2\pi}$

$V = 0 + 2\pi \times 1000 = 2000\pi\ m^{3}$

7 0

3 years ago

Which are the solutions of the equation x^2+4x+4=25

pav-90 [236]

I think the answer is 3,-7

8 0

3 years ago

A Pew Internet poll asked cell phone owners about how they used their cell phones. One question asked whether or not during the

EastWind [94]

Answer:

a) $\hat p=\frac{471}{1024}=0.460$

The standard error is given by:

$SE= \sqrt{\frac{\hat p(1-\hat p)}{n}}=\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0156$

And the margin of error is given by:

$ME=z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}=1.96\sqrt{\frac{0.460(1-0.460)}{1024}}=0.0305$

b) The 99% confidence interval would be given by (0.429;0.491)

Step-by-step explanation:

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

Data given and notation

n=1024 represent the random sample taken

X=471 represent the people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

$\hat p=\frac{471}{1024}=0.460$ estimated proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

p= population proportion of people responded that they had used their cell phone while in a store within the last 30 days to call a friend or family member for advice about a purchase they were considering

Part a

The confidence interval would be given by this formula

$\hat p \pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}$

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$ , with that value we can find the quantile required for the interval in the normal standard distribution.