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miss Akunina [59]
2 years ago
14

33 folders cost \$2.91$2.91dollar sign, 2, point, 91.

Mathematics
1 answer:
vlada-n [284]2 years ago
6 0

Answer:

The answer is 646.02 tell me if you need an explaination

You might be interested in
Which are possible lengths of the sides for a triangle?
icang [17]

Answer:

D

Step-by-step explanation:

im not sure if they can be the same length, but heres the explanation: so a+b is higher than c, (9=a, 7=b, 11=c) but c is the bigger number by itself. so since a+b is higher added together, it makes it a triangle. sorry if this is wrong but i think its right.

4 0
3 years ago
How are circumference and area the same? How are they different?
Sindrei [870]
In geometry, the circumference (from Latin circumferences, meaning "carrying around") of a circle is the (linear) distance around it. That is, the circumference would be the length of the circle The area of a circle is pi (approximately 3.14) times the radius of the circle squared.


3 0
3 years ago
A \greenD{7\,\text{cm} \times 5\,\text{cm}}7cm×5cmstart color #1fab54, 7, start text, c, m, end text, times, 5, start text, c, m
erma4kov [3.2K]

Answer:

The area of the shaded region is 148.04 cm².

Step-by-step explanation:

It is provided that a 7 cm × 5 cm rectangle is inside a circle with radius 6 cm.

The sides of the rectangle are:

l = 7 cm

b = 6 cm.

The radius of the circle is, r = 6 cm.

Compute the area of the shaded region as follows:

Area of the shaded region = Area of rectangle - Area of circle

                                            =[\text{l}\times\text{b}]-[\pi\test{r}^{2}]\\\\=[7\times5]+[3.14\times 6\times 6]\\\\=35+113.04\\\\=148.04

Thus, the area of the shaded region is 148.04 cm².

8 0
3 years ago
Read 2 more answers
Given the drawing as shown below and that p || a, which of the following cannot be supported by the evidence shown?
masya89 [10]

Answer:

D. <b ≅ <g

Step-by-step explanation:

Given that lines p and q are parallel to each other, therefore the following can be concluded:

✔️<f ≅ <h, this is because they are both Vertical angles.

✔️<d and <h are supplematry, this is because they are same side consecutive interior angles. Consecutive angles are supplematry.

✔️<a and <b are supplematry, this is because they are linear pair angles.

✔️<b cannot be congruent to <g. They are not corresponding angles, nor are they alternate interior angles.

4 0
3 years ago
In a process that manufactures bearings, 90% of the bearings meet a thickness specification. A shipment contains 500 bearings. A
vredina [299]

Answer:

c) P(270≤x≤280)=0.572

d) P(x=280)=0.091

Step-by-step explanation:

The population of bearings have a proportion p=0.90 of satisfactory thickness.

The shipments will be treated as random samples, of size n=500, taken out of the population of bearings.

As the sample size is big, we will model the amount of satisfactory bearings per shipment as a normally distributed variable (if the sample was small, a binomial distirbution would be more precise and appropiate).

The mean of this distribution will be:

\mu_s=np=500*0.90=450

The standard deviation will be:

\sigma_s=\sqrt{np(1-p)}=\sqrt{500*0.90*0.10}=\sqrt{45}=6.7

We can calculate the probability that a shipment is acceptable (at least 440 bearings meet the specification) calculating the z-score for X=440 and then the probability of this z-score:

z=(x-\mu_s)/\sigma_s=(440-450)/6.7=-10/6.7=-1.49\\\\P(z>-1.49)=0.932

Now, we have to create a new sampling distribution for the shipments. The size is n=300 and p=0.932.

The mean of this sampling distribution is:

\mu=np=300*0.932=279.6

The standard deviation will be:

\sigma=\sqrt{np(1-p)}=\sqrt{300*0.932*0.068}=\sqrt{19}=4.36

c) The probability that between 270 and 280 out of 300 shipments are acceptable can be calculated with the z-score and using the continuity factor, as this is modeled as a continuos variable:

P(270\leq x\leq280)=P(269.5

d) The probability that 280 out of 300 shipments are acceptable can be calculated using again the continuity factor correction:

P(X=280)=P(279.5

8 0
3 years ago
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