All categories

2 years ago

Can someone help me with these two math problems. ( Will Mark Brainliest).

Mathematics

1 answer:

SOVA2 [1]2 years ago

5 0

Answer:

A 83 cups

B 16 cups.

Step-by-step explanation:

A) the volume of the cone = 1/3 pi r^2 h

= 1/3 pi 2^2 * 6

= 8pi

Number of cups needed = 2000/3 pi / 8 pi

= 83 cups to the nearest cup.

B) Volume of this cup is 1/3 pi 4^2 * 8 = 128/3 pi

Number of cups required = 2000/ 3 * 3 / 128

= 15. 6 cups.

You might be interested in

_(9)=(2(1-2(2)^(9)))/(1-2(2))<br><br><br>PLEASE HELP ME OUT.

USPshnik [31]

Answer:

the answer = um ok

Step-by-step explanation:

4 0

2 years ago

What is the value of x in the diagram?

PilotLPTM [1.2K]

Answer:

Step-by-step explanation:

They are vertical angles; so angles are congruent

*<em>Vertical angles are always congruent*</em>

<em>Congruent means equal or same.</em>

1) Since they are congruent, you form an equation.

2) Solve by subtracting 15 from both sides.

3) Solve by dividing 2 from both sides

4) Simplify

x = 65

7 0

2 years ago

8 less than a number n is less than 11

iragen [17]

In mathematical form, it would be:

n-8 < 11

add 8 from both sides,

n-8+8 < 11+8

n < 19

So, n value must be less than 19

Hope this helps!

8 0

2 years ago

Rockwell hardness of pins of a certain type is known to have a mean value of 50 and a standard deviation of 1.2.a. If the distri

zalisa [80]

Answer:

$P(\= X \ge 51 ) =0.0062$

$P(\= X \ge 51 ) = 0$

Step-by-step explanation:

From the question we are told that

The mean value is $\mu = 50$

The standard deviation is $\sigma = 1.2$

Considering question a

The sample size is n = 9

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{9} }$

=> $\sigma_x = 0.4$

Generally the probability that the sample mean hardness for a random sample of 9 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_{x}} \ge \frac{51 - 50 }{0.4 } )$

$\frac{\= X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ \= X )$

$P(\= X \ge 51 ) = P( Z \ge 2.5 )$

=> $P(\= X \ge 51 ) =1- P( Z < 2.5 )$

From the z table the area under the normal curve to the left corresponding to 2.5 is

$P( Z < 2.5 ) = 0.99379$

=> $P(\= X \ge 51 ) =1-0.99379$

=> $P(\= X \ge 51 ) =0.0062$

Considering question b

The sample size is n = 40

Generally the standard error of the mean is mathematically represented as

$\sigma_x = \frac{\sigma }{\sqrt{n} }$

=> $\sigma_x = \frac{ 1.2 }{\sqrt{40} }$

=> $\sigma_x = 0.1897$

Generally the (approximate) probability that the sample mean hardness for a random sample of 40 pins is at least 51 is mathematically represented as

$P(\= X \ge 51 ) = P( \frac{\= X - \mu }{\sigma_x} \ge \frac{51 - 50 }{0.1897 } )$

=> $P(\= X \ge 51 ) = P(Z \ge 5.2715 )$

=> $P(\= X \ge 51 ) = 1- P(Z < 5.2715 )$

From the z table the area under the normal curve to the left corresponding to 5.2715 and

=> $P(Z < 5.2715 ) = 1$

$P(\= X \ge 51 ) = 1- 1$

=> $P(\= X \ge 51 ) = 0$

5 0

2 years ago

an art teacher has 1 1/2 pounds of red clay and 3/4 pound of yellow clay.The teacher mixes the red clay and yellow clay together

Nataly [62]

Well, seeing as how she mixed the clay, she has a total of 2 2/8 pounds clay (unsimplified). Although we don't know how many students she has in one class, with simple division of this number we can figure out how many student can finish the project. Removing the 2/8, we can already tell 2 students can do it, so that's a start. Now to divide the 2 pounds into 8ths. Each pound is enough for 8 students, so the 2 pounds totals to 16. Now add the extra 2 students, it's 18 total students. Hope I explained it well enough!

7 0

2 years ago

Can someone help me with these two math problems. ( Will Mark Brainliest). ​

Can someone help me with these two math problems. ( Will Mark Brainliest).