1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Temka [501]
3 years ago
11

Jammal makes a cut through a block of florist's foam, as shown. What are the dimensions of the exposed cross section?

Mathematics
1 answer:
Ahat [919]3 years ago
8 0

Answer:

C. 3 in x 6 in

Step-by-step explanation:

Jammal cuts the block in a straight line parallel to one side... so the section revealed when he finishes his cut will be identical as the parallel side to which the cut is done.

We know the the left side of the prism on the image is 3 inches wide and 6 inches high... so that will also be the dimensions of the exposed cross section.

The answer is then 3 inches y 6 inches. The thickness of the block (5 inches) has no impact on the exposed area of the cross-section.

You might be interested in
What are the rise and run of the line whose equation is y=15x ?
butalik [34]

y = mx , where m is a slope,

m= rise/run

We have y = 15x, so m =15 and  rise/run =15, then

rise = 15, run = 1

5 0
3 years ago
subtract the product of 3 and 4 from 36. then divide the quotient of 9 and 3, subtracted from fifteen
Butoxors [25]
The first one is 24, and the second one is 12
4 0
3 years ago
Read 2 more answers
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
Marina86 [1]

Answer:

(a) 0.94

(b) 0.20

(c) 90.53%

Step-by-step explanation:

From a population (Bernoulli population), 90% of the bearings meet a thickness specification, let p_1 be the probability that a bearing meets the specification.

So, p_1=0.9

Sample size, n_1=500, is large.

Let X represent the number of acceptable bearing.

Convert this to a normal distribution,

Mean: \mu_1=n_1p_1=500\times0.9=450

Variance: \sigma_1^2=n_1p_1(1-p_1)=500\times0.9\times0.1=45

\Rightarrow \sigma_1 =\sqrt{45}=6.71

(a) A shipment is acceptable if at least 440 of the 500 bearings meet the specification.

So, X\geq 440.

Here, 440 is included, so, by using the continuity correction, take x=439.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{339.5-450}{6.71}=-1.56.

So, the probability that a given shipment is acceptable is

P(z\geq-1.56)=\int_{-1.56}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}}=0.94062

Hence,  the probability that a given shipment is acceptable is 0.94.

(b) We have the probability of acceptability of one shipment 0.94, which is same for each shipment, so here the number of shipments is a Binomial population.

Denote the probability od acceptance of a shipment by p_2.

p_2=0.94

The total number of shipment, i.e sample size, n_2= 300

Here, the sample size is sufficiently large to approximate it as a normal distribution, for which mean, \mu_2, and variance, \sigma_2^2.

Mean: \mu_2=n_2p_2=300\times0.94=282

Variance: \sigma_2^2=n_2p_2(1-p_2)=300\times0.94(1-0.94)=16.92

\Rightarrow \sigma_2=\sqrt(16.92}=4.11.

In this case, X>285, so, by using the continuity correction, take x=285.5 to compute z score for the normal distribution.

z=\frac{x-\mu}{\sigma}=\frac{285.5-282}{4.11}=0.85.

So, the probability that a given shipment is acceptable is

P(z\geq0.85)=\int_{0.85}^{\infty}\frac{1}{\sqrt{2\pi}}e^{\frac{-z^2}{2}=0.1977

Hence,  the probability that a given shipment is acceptable is 0.20.

(c) For the acceptance of 99% shipment of in the total shipment of 300 (sample size).

The area right to the z-score=0.99

and the area left to the z-score is 1-0.99=0.001.

For this value, the value of z-score is -3.09 (from the z-score table)

Let, \alpha be the required probability of acceptance of one shipment.

So,

-3.09=\frac{285.5-300\alpha}{\sqrt{300 \alpha(1-\alpha)}}

On solving

\alpha= 0.977896

Again, the probability of acceptance of one shipment, \alpha, depends on the probability of meeting the thickness specification of one bearing.

For this case,

The area right to the z-score=0.97790

and the area left to the z-score is 1-0.97790=0.0221.

The value of z-score is -2.01 (from the z-score table)

Let p be the probability that one bearing meets the specification. So

-2.01=\frac{439.5-500  p}{\sqrt{500 p(1-p)}}

On solving

p=0.9053

Hence, 90.53% of the bearings meet a thickness specification so that 99% of the shipments are acceptable.

8 0
3 years ago
The side length of a square is 4 inches. Find the perimeter of the square. HELP PLEASE!
Elan Coil [88]

Answer:

16 in

Step-by-step explanation:

4 +4+4+4 = 16

6 0
2 years ago
Read 2 more answers
1/2 of 3oz, 3/4 of 3oz and 1/3 of 2oz, how many ounces of food was consumed
Alex Ar [27]
4.42 ounces of food was consumed. 1/2 of 3 oz equals 1.5 oz, 3/4 of 3 oz equals 2.25 oz and 1/3 of 2 oz equals .67 and when you add those up you get 4.42.
3 0
3 years ago
Other questions:
  • ​ ​ A hotel has 18 floors. The hotel owner believes the number 13 is unlucky. The first 12 floors are numbered from 1 to 12. Flo
    12·1 answer
  • Find the factors of 582.
    13·1 answer
  • HELPPPP ME PLEASEEEE!!!!
    7·2 answers
  • X=y-3 4x+2y=2? What is the answer
    6·1 answer
  • 700,000+9,000+60+7 what is the standard form
    8·2 answers
  • What’s the slope? I need help ASAP;(
    9·2 answers
  • There is a bag filled with 5 blue and 6 red marbles.
    9·2 answers
  • B||C find the value of y 71 (2y+34)
    14·1 answer
  • Solve for the missing variable.<br> 1.)<br> 80/120 =2/x
    8·1 answer
  • The data set below shows the admission price​ (in dollars) for​ one-day tickets to 10 theme parks in the United States. What is
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!