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2 years ago

What is the square root of 4

Mathematics

1 answer:

Masja [62]2 years ago

8 0

Answer:

√4 = 2

Because 2 × 2 = 4

If you want to get square root of any number find the number which you can multiply two times to give you that certain number

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there is a walking trail at the park. four laps around the trail is a distance of 1 mile. how many laps does it take to walk 3/4

dybincka [34]

1 ÷ 4 1/4 so one lap is a quarter of a mile meaning 3 laps is 3/4 of a mile

4 0

3 years ago

Eva wants to make two pieces of pottery.She needs 3/5 pound of clay for one piece and 7/10 pound of clay for the other piece.She

marta [7]

Answer:

a) Eva will need 2 bags of clay to make both pieces of pottery.

b) She will have leftover $\frac{11}{10}$ pounds of clay.

Step-by-step explanation:

a) We have to find the total amount of clay, as she needs $\frac{3}{5}pound=P_{1}$ and $\frac{7}{10}pound=P_{2}$ , so $P_{1}+P_{2}= \frac{3}{5} +\frac{7}{10} =\frac{6+7}{10}=\frac{13}{10}pounds$ , is the total amount of clay. Then as we Know that one bag of clay has $\frac{4}{5} =\frac{8}{10} pounds$ , It isn´t enough, but with two bags $2*\frac{8}{10}=\frac{16}{10} pounds$ , It is enough clay.

b) She will have leftover $\frac{16}{10} -\frac{13}{10} =\frac{3}{10}pounds$ of the second bag, then we add the third one, so $\frac{3}{10} +\frac{8}{10} =\frac{11}{10}pounds$ are leftover.

6 0

2 years ago

Is math important in real life?

Dmitrij [34]

Yes- though we may say "When will we ever use this?" ever so often in class, the reality is that we use mathematics in everyday life. From simple addition, to factoring, to finding the angles of various components to a building, math is always being used in real life.

6 0

3 years ago

Read 2 more answers

The distribution of heights for adult men in a certain population is approximately normal with mean 70 inches and standard devia

KengaRu [80]

Answer:

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12]

Step-by-step explanation:

I beleive those options corresponds to another question, i will ignore them. We want to know an interval in which the probability that a height falls there is 0.8.

In such interval, the probability that a value is higher than the right end of the interval is (1-0.8)/2 = 0.1

If X is the distribuition of heights, then we want z such that P(X > z) = 0.1. We will take W, the standarization of X, wth distribution N(0,1)

$W = \frac{X-\mu}{\sigma} = \frac{X-70}{4}$

The values of the cumulative distribution function of W, denoted by $\phi$ , can be found in the attached file. Lets call $y = \frac{z-70}{4}$ . We have

$0.1 = P(X > z) = P(\frac{X-70}{4} > \frac{z-70}{4}) = P(W > y) = 1-\phi(y)$

Thus

$\phi(y) = 1-0.1 = 0.9$

by looking at the table, we find that y = 1.28, therefore

$\frac{z-70}{4} = 1.28\\z = 1.28*4+70 = 75.12$

The other end of the interval is the symmetrical of 75.12 respect to 70, hence it is 70- (75.12-70) = 64.88.

The interval (meassured in Inches) that represent the middle 80% of the heights is [64.88, 75.12] .

Download pdf

8 0

2 years ago

Find an equation for the nth term of the arithmetic sequence.

Irina18 [472]

Answer:

The equation of the nth term is an = -621 + 42n

Step-by-step explanation:

* Lets revise the arithmetic sequence

- There is a constant difference between each two consecutive numbers

- Ex:

# 2 , 5 , 8 , 11 , ……………………….

# 5 , 10 , 15 , 20 , …………………………

# 12 , 10 , 8 , 6 , ……………………………

* General term (nth term) of an Arithmetic sequence:

- U1 = a , U2 = a + d , U3 = a + 2d , U4 = a + 3d , U5 = a + 4d

- Un = a + (n – 1)d, where a is the first term , d is the difference

between each two consecutive terms

, n is the position of the term

* Lets solve the problem

∵ an = a + (n - 1)d

∴ a14 = a + (14 - 1)d

∴ a14 = a + 13d

∵ a14 = -33

∴ a + 13d = -33 ⇒ (1)

- Similar we can find another equation from a15

∵ a15 = a + (15 - 1)d

∴ a15 = a + 14d

∵ a15 = 9

∴ a + 14d = 9 ⇒ (2)

- We will solve equations (1) and (2) to find a and d

* Lets subtract equation (2) from equation (1)

∴ (a - a) + (13 - 14)d = (-33 - 9)

∴ -d = -42 ⇒ × both sides by -1

∴ d = 42

- Substitute this value of d in equation (1) or (2)

∵ a + 13d = -33

∵ d = 42

∴ a + 13(42) = -33

∴ a + 546 = -33 ⇒ subtract 546 from both sides

∴ a = -579

* Now lets write the equation of the nth term

∵ an = a + (n - 1)d

∵ a = -579 and d = 42

∴ an = -579 + (n - 1) 42 ⇒ open the bracket

∴ an = -579 + 42n - 42

∴ an = -621 + 42n

* The equation of the nth term is an = -621 + 42n

6 0

3 years ago