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3 years ago

PLEASE I NEED HELP, ILY ILL GIVE 30 POINTS

Mathematics

1 answer:

Novosadov [1.4K]3 years ago

7 0

Answer:

Question 1:

The angles are presented here using Autocad desktop application

The two column proof is given as follows;

Statement ${}$ Reason

S1. Line m is parallel to line n ${}$ R1. Given

S2. ∠1 ≅ ∠2 ${}$ R2. Vertically opposite angles

S3. m∠1 ≅ m∠2 ${}$ R3. Definition of congruency

S4. ∠2 and ∠3 form a linear pear ${}$ R4. Definition of a linear pair

S5. ∠2 is supplementary to ∠3 ${}$ R5. Linear pair angles are supplementary

S6. m∠2 + m∠3 = 180° ${}$ Definition of supplementary angles

S7. m∠1 + m∠3 = 180° ${}$ Substitution Property of Equality

S8. ∠1 is supplementary to ∠3 ${}$ Definition of supplementary angles

Question 2:

(a) The property of a square that is also a property of a rectangle is that all the interior angles of both a square and a rectangle equal

(b) The property of a square that is not necessarily a property of all rectangles is that the sides of a square are all equal, while only the length of the opposite sides of a rectangle are equal

(d) A property of a rhombi that is not necessarily a property of all parallelogram is that the diagonals of a rhombi are perpendicular

(e) A property that applies to all parallelogram is that the opposite sides of all parallelogram are equal

Step-by-step explanation:

You might be interested in

* Ms. Quintero has visited 16 of the 50 states. What percent of the states has she visited? KEY Part (is) Whole(of) % 100 Whole

riadik2000 [5.3K]

Answer:

B. 32%

Step-by-step explanation:

16/50 = 32% lol

goodluck!

5 0

2 years ago

A manufacturer of golf equipment wishes to estimate the number of left-handed golfers. A previous study indicates that the propo

never [62]

Answer:

A sample of 997 is needed.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence level of $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

The margin of error is of:

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

A previous study indicates that the proportion of left-handed golfers is 8%.

This means that $\pi = 0.08$

98% confidence level

So $\alpha = 0.02$ , z is the value of Z that has a p-value of $1 - \frac{0.02}{2} = 0.99$ , so $Z = 2.327$ .

How large a sample is needed in order to be 98% confident that the sample proportion will not differ from the true proportion by more than 2%?

This is n for which M = 0.02. So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 2.327\sqrt{\frac{0.08*0.92}{n}}$

$0.02\sqrt{n} = 2.327\sqrt{0.08*0.92}$

$\sqrt{n} = \frac{2.327\sqrt{0.08*0.92}}{0.02}$

$(\sqrt{n})^2 = (\frac{2.327\sqrt{0.08*0.92}}{0.02})^2$

$n = 996.3$

Rounding up:

A sample of 997 is needed.

3 0

3 years ago

Find all of the zeros of the function:

IceJOKER [234]

Answer:Roots: 3;2;-1

Step-by-step explanation: i use horner's scheme for approximating the roots of polynomials

6 0

3 years ago

I asked this earlier and got no answer. Will give brainliest

Virty [35]

The area of a square is

s • s

We can also write this as

s^2

So, for any side length “s”, we can make a function, A(s), such that

A(s) = s^2

Now that we have a quadratic equation for the area of a square, let’s go ahead and solve for the side lengths of a square with a given area. In this case, 225 in^2

225 = s^2

Therefore,

s = sqrt(225)

s = 15

So, the dimensions are 15 x 15 in

6 0

3 years ago

James has 14{,}500\text { g}14,500 g14, comma, 500, start a text, space, g, end text of sand in his sandbox. He brings home anot

mariarad [96]

Question:

James has 14,500g of sand in his sandbox. He brings home another 7,400g of sand from the beach to add to his sandbox. How many kilograms of sand does James have in his sandbox now?

Answer:

The quantity of sandbox James has is 21.9kg

Step-by-step explanation:

Given

Initial quantity of sandbox = 14,500 g

Additional quantity = 7,400 g

Required

Total quantity of sandbox in kg

Tp get the total quantity of sandbox, we simply add in the initial and additional quantity of sandbox together;

This is represented mathematically as follows;

Total = Initial Quantity + Additional Quantity

Total = 14,500 g + 7,400 g

Total = 21,900 g

But the question says the answer should be in kg

1000 g is equivalent to 1 kg

21,900 g will be equivalent to $\frac{21,900}{1,000} kg$

$21,900g = \frac{21,900}{1,000} kg$

$21,900g = 21.9 kg$

Hence, the quantity of sandbox James has is 21.9kg

7 0

3 years ago