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2 years ago

Help me please thanks

Mathematics

1 answer:

Zarrin [17]2 years ago

8 0

FORMULA:

Pythagorean property —: h² = b² + p², where h = hypotenuse, b = base, p = perpendicular.

ANSWER:

By pythagorean property,

x² = 25² + 100²
x² = 625 + 10000
x² = 10625
x = √10625
x = 103 cm rounded.

Hence, the length of x is 103 cm.

You might be interested in

A carton can hold 1,000 unit cubes that measure 1 inch by inch.describe the dimensions of the carton using unit cubes

Romashka-Z-Leto [24]

You've told us the volume of the carton.
That doesn't tell us the dimensions.

If the unit cubes have to exactly fit the dimensions,
then all we know is that the dimensions are integers,
and their product is 1,000.

The carton could be

1 by 1 by 1,000
1 by 2 by 500
1 by 4 by 250
1 by 5 by 200
1 by 8 by 125
1 by 10 by 100
1 by 20 by 50
1 by 25 by 40

2 by 2 by 250
2 by 5 by 100
2 by 10 by 50
.
.
and several other possible sets.

6 0

3 years ago

An interior automotive supplier places several electrical wires in a harness.Apull test measures the force required to pull spli

oksano4ka [1.4K]

Answer:

a) For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assumin Normal distribution

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical dsitribution

The result is on the figure attached.

b) For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

Step-by-step explanation:

For this case we have the following data:

28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4

The quantile-quantile or q-q plot is a graphical procedure in order to check the validity of a distributional assumption for a data set. We just need to calculate "the theoretically expected value for each data point based on the distribution in question".

If the values are asusted to the assumed distribution, we will see that "the points on the q-q plot will fall approximately on a straight line"

For this case our distribution assumed is normal.

Part a

For this case we can use the following R code to construct the qq plot

> data<-c(28.8, 24.4, 30.1, 25.6, 26.4, 23.9, 22.1, 22.5, 27.6, 28.1, 20.8, 27.7, 24.4, 25.1, 24.6, 26.3, 28.2, 22.2, 26.3, 24.4)

# The above line is in order to store the data in a vector

> qqnorm(data, pch = 1, frame = FALSE)

# The line above is in order to calculate the quantiles from the data assuming Normal distribution (0,1)

> qqline(data, col = "steelblue", lwd = 2)

# The line above is in order to put a line for the theoretical distribution

The result is on the figure attached.

Part b

For this case as we can see on the figure attached the calculated quantiles are not far from the theorical quantiles given byt the straaigth blue line so then we can conclude that the distribution seems to be normal.

4 0

3 years ago

PLEASE HELP DUE IN 30 MINUTES

Scilla [17]

Answer: B

Step-by-step explanation: Length x Width = Area

8 feet x width = 96 sq ft

8w = 96

5 0

3 years ago

Read 2 more answers

Prove each of the following statements below using one of the proof techniques and state the proof strategy you use.

pochemuha

Answer:

See below

Step-by-step explanation:

a) Direct proof: Let m be an odd integer and n be an even integer. Then, there exist integers k,j such that m=2k+1 and n=2j. Then mn=(2k+1)(2j)=2r, where r=j(2k+1) is an integer. Thus, mn is even.

b) Proof by counterpositive: Suppose that m is not even and n is not even. Then m is odd and n is odd, that is, m=2k+1 and n=2j+1 for some integers k,j. Thus, mn=4kj+2k+2j+1=2(kj+k+j)+1=2r+1, where r=kj+k+j is an integer. Hence mn is odd, i.e, mn is not even. We have proven the counterpositive.

c) Proof by contradiction: suppose that rp is NOT irrational, then rp=m/n for some integers m,n, n≠. Since r is a non zero rational number, r=a/b for some non-zero integers a,b. Then p=rp/r=rp(b/a)=(m/n)(b/a)=mb/na. Now n,a are non zero integers, thus na is a non zero integer. Additionally, mb is an integer. Therefore p is rational which is contradicts that p is irrational. Hence np is irrational.

d) Proof by cases: We can verify this directly with all the possible orderings for a,b,c. There are six cases:

a≥b≥c, a≥c≥b, b≥a≥c, b≥c≥a, c≥b≥a, c≥a≥b

Writing the details for each one is a bit long. I will give you an example for one case: suppose that c≥b≥a then max(a, max(b,c))=max(a,c)=c. On the other hand, max(max(a, b),c)=max(b,c)=c, hence the statement is true in this case.

e) Direct proof: write a=m/n and b=p/q, with m,q integers and n,q nonnegative integers. Then ab=mp/nq. mp is an integer, and nq is a non negative integer. Hence ab is rational.

f) Direct proof. By part c), √2/n is irrational for all natural numbers n. Furthermore, a is rational, then a+√2/n is irrational. Take n large enough in such a way that b-a>√2/n (b-a>0 so it is possible). Then a+√2/n is between a and b.

g) Direct proof: write m+n=2k and n+p=2j for some integers k,j. Add these equations to get m+2n+p=2k+2j. Then m+p=2k+2j-2n=2(k+j-n)=2s for some integer s=k+j-n. Thus m+p is even.

7 0

3 years ago

Kelly uses a building block to create the top of a tower. The front face of the building block is shown.

max2010maxim [7]

What you need to do is do the formula of Area=1/2×Height (Base B +Base A) Base B is bottom Base A is the top then you divide by 2 In this case 1/2 ×8/1 ×(12+15)=108

3 0

2 years ago

Help me please thanks​

Help me please thanks