Ask question

Ask question

All categories

3 years ago

5

The formula for finding the volume of a cone is V = 1/3πr2h. The volume of a cone is 300 cm3 and the height of the cone is 10 cm

. What is the approximate radius of the cone?
A) 3 cm
B) 5 cm
C) 9 cm
D) 28 cm

1 answer:

stepladder [879]3 years ago

5 0

The closest approximate answer would be B. 5cm

You might be interested in

PLSS HELP IF YOU TURLY KNOW THISS

coldgirl [10]

Answer:

x = 7

Step-by-step explanation:

$7x = 49$

$7x \div 7 = 49 \div 7$

$x = 49 \div 7 = 7$

$\boxed{\green{x = 7}}$

3 0

3 years ago

Read 2 more answers

The answer to a,b and c

GuDViN [60]

Answer:

16/45; 4/45; 1125

Step-by-step explanation:

Since you know that 4/9 and 1/5 are used, make them common denominators. So 4/9 would turn into 20/45 and 1/5 would be 9/45. It results in 29/45, but that is the amount spent, so you would subtract 45-29/45, resulting in 16/45. For B, you know that she spent 3/4 of the savings, so you know that 1/4 is not being spent. So use 1/4 multiplied by 16/45 from part A to get 4/45. For C, you know that 4/45 is $100, so I multiplied 45x100 to get 4500 and then divided by 4 to get 4500/4.

8 0

3 years ago

How can I show money as a fraction?

abruzzese [7]

You can show money as a fraction by taking the amount of cents, and putting it over 100.

Ex. $1.47 = 147 cents
147 cents = 147/100 in fraction form.

4 0

3 years ago

Suppose the heights of a population of people are normally distributed with a mean of 65.5 inches and a standard deviation of 2.

kupik [55]

Answer:

a) $P(64.2$

b) 69.764

Step-by-step explanation:

1) Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

2) Part a

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(65.5,2.6)$

Where $\mu=65.5$ and $\sigma=2.6$

We are interested on this probability

$P(64.2$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

$z=\frac{x-\mu}{\sigma}$

If we apply this formula to our probability we got this:

$P(64.2$

And we can find this probability on this way:

$P(-0.50$

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

$P(-0.50$

3) Part b

For this part we want to find a value a, such that we satisfy this condition:

$P(X>a)=0.05$ (a)

$P(X$ (b)

Both conditions are equivalent on this case. We can use the z score again in order to find the value a.

As we can see on the figure attached the z value that satisfy the condition with 0.95 of the area on the left and 0.05 of the area on the right it's z=1.64. On this case P(Z<1.64)=0.95 and P(z>1.64)=0.05

If we use condition (b) from previous we have this:

$P(X$

$P(z$

But we know which value of z satisfy the previous equation so then we can do this:

$z=1.64$

And if we solve for a we got

$a=65.5 +1.64*2.6=69.764$

So the value of height that separates the bottom 95% of data from the top 5% is 69.764.

5 0

3 years ago

What are the real and complex solutions of the polynomial equation? x^4-41x^2=-400. *work needed*

pickupchik [31]

<h3>Answer: x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

Step-by-step explanation:

Given the polynomial equation $x^4-41x^2=-400$ .

Adding 400 on both sides to get rid 400 from right side and set 0 on right side, we get

$x^4-41x^2+400=-400+400$ .

$x^4-41x^2+400=0$ .

Factoring by product sum rule.

We need product of 400 and sum upto -41.

We can see that 400 = -25 × -16 = 400 and -25-16 = -41.

Therefore,

$x^4-25x^2-16x^2+400=0$

Making it into two groups, we get

$(x^4-25x^2)+(-16x^2+400)=0$

Factoring out GCF of each group, we get

$x^2(x^2-25)-16(x^2-25)=0$

$(x^2-25)(x^2-16) =0$

Factoring out $(x^2-25)$ and $(x^2-16)$ separately by difference of the squares identity $a^2-b^2=(a-b)(a+b)$ , we get

$(x^2-25) = x^2-5^2= (x-5)(x+5)$ and

$x^2-16 = x^2-4^2 =(x-4)(x+4)$ .

Therefore,

$(x-5)(x+5)(x-4)(x+4) =0$

Applying zero product rule,

x-5=0

x+5=0

x-4=0 and

x+4=0.

Therefore,

<h3>x = -5, -4, 4 and 5.</h3><h3>All four zeros are real solutions.</h3>

6 0

3 years ago