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liraira [26]
3 years ago
8

Following an experiment, Jamie estimates that his probability of making a three-pointer with a basketball is 0.8. If he shoots 1

60 attempts from the three-point line, how many times can he expect to make a three-point basket?
Mathematics
2 answers:
dem82 [27]3 years ago
6 0
Multiply the number of attempts with the probability,
160 x 0.8 = 128 attempts made.

Hope this helps.
NeTakaya3 years ago
6 0

Answer:

C) 128

Step-by-step explanation:

This is on plato

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Help please! Complete the proof!
Dmitry [639]
Since CM is perpendicular to AB, it follows that ∠1 and <span>∠2 are 90 degrees.  Since they both have 90 degree angles, that must mean they are both right triangles.

And since </span>∠3 = ∠4 and ∠1 = ∠2, then it follows that ∠A = <span>∠B.  (You can show this by showing that they must add up to 180 degrees.)

So since both right triangles have 3 congruent angles to each, then that makes them similar by AAA (angle angle angle).</span>
5 0
3 years ago
Find the polynomial of minimum degree, with real coefficients, zeros at
drek231 [11]

Answer:

\huge\boxed{p(x)=4x^3-20x^2+4x+300}

Step-by-step explanation:

\text{If}\ x=4\pm3i\ \text{and}\ x=-3\ \text{are the zeros of a polynomial, then it has  a form:}\\\\p(x)=\bigg(x-(4-3i)\bigg)\bigg(x-(4+3i)\bigg)\bigg(x-(-3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x-4+3i)(x-4-3i)(x+3)\bigg(r(x)\bigg)\\\\p(x)=\underbrace{\bigg((x-4)+3i\bigg)\bigg((x-4)-3i\bigg)}_{\text{use}\ (a+b)(a-b)=a^2-b^2}(x+3)\bigg(r(x)\bigg)\\\\p(x)=\bigg((x-4)^2-(3i)^2\bigg)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ (a-b)^2=a^2-2ab+b^2

p(x)=(x^2-2(x)(4)+4^2-3^2i^2)(x+3)\bigg(r(x)\bigg)\qquad\text{use}\ i^2=-1\\\\p(x)=(x^2-8x+16-9(-1))(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+16+9)(x+3)\bigg(r(x)\bigg)\\\\p(x)=(x^2-8x+25)(x+3)\bigg(r(x)\bigg)\qquad\text{use FOIL}:\ (a+b)(c+d)=ac+ad+bc+bd\\\\p(x)=\bigg((x^2)(x)+(x^2)(3)+(-8x)(x)+(-8x)(3)+(25)(x)+(25)(3)\bigg)\bigg(r(x)\bigg)\\\\p(x)=(x^3+3x^2-8x^2-24x+25x+75)\bigg(r(x)\bigg)\qquad\text{combine like terms}\\\\p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)

\text{The y-intercept is at 300}.\\\\\text{For}\ w(x)=a_nx^n+a_{n-1}x^{n-1}+a_{n-2}x^{n-2}+...+a_1x+a_0\\\\\text{y-intercept is}\ a_0\\\\\text{Therefore for}\ p(x)=(x^3-5x^2+x+75)\bigg(r(x)\bigg)\\\\\text{y-intercet is}\ 75\bigg(r(x)\bigg)\\\\75\bigg(r(x)\bigg)=300\qquad\text{divide both sides by 75}\\\\r(x)=4\\\\\text{Finally:}\\\\p(x)=(x^3-5x^2+x+75)(4)\qquad\text{use the distributive property}\\\\p(x)=(x^3)(4)+(-5x^2)(4)+(x)(4)+(75)(4)\\\\p(x)=4x^3-20x^2+4x+300

7 0
3 years ago
Find the balance at the end of 4 years if $10,000 is deposited at a rate of 1.5% simple interest.
Minchanka [31]

Answer:

Principal amount (P) = $10,000

Rate (R) = 1.5%

Time (T) = 4 years

Simple interest, I = P X R X T / 100

= 10000 X 1.5 X 4 /100

= 60000 / 100

= $600

Therefore, Balance = P + I

= 10000 + 600

= $10600

3 0
3 years ago
Quiz 3
erica [24]

Complete question :

The data below represents the number of customers at each Slurpee Sam's Spaghetti Shop. 24, 25, 29, 30, 31, 31, 32, 34, 34. Which box plot correctly summarizes the data?

Answer:

B

Step-by-step explanation:

BOX plots provide information about :

Minimum, Lower quartile, Median, Upper quartile and maximum value of a dataset.

Given the dataset :

24, 25, 29, 30, 31, 31, 32, 34, 34

Minimum value = 24 (Lowest value in the dataset)

First quartile (Q1) = 1/4(n+1)th term

Q1 = 1/4 * 10 = 10/4 = 2.5th term = (25+29)/2 = 27

Second quartile ; median (Q2) = 1/2(n+1)th term

Q2 = 1/2(n+1)th term = 1/2(10) = 5th = 31

Third quartile (Q3) = 3/4(n+1)th term

Q3 = 3/4(n+1)th term = 3/4(10) = 30/4 = 7.5th term = (7th +8th) term = (32 + 34) / 2 = 66/2 = 33

Maximum value = 34 (Highest value in the dataset)

First dot = minimum value

Last dot = maximum value

First line of box = QI

mid line = median

End line of box = Q3

4 0
2 years ago
How many 4 digit multiples can not be divided by 5
Advocard [28]

Answer:

1800

Step-by-step explanation:

Let us solve this question by Modular Airthmatic.

Let the lowest 4 digit number be X.

X= 1 (mod 5)

X =2 (mod 5)

X=3 (mod 5)

X =4 (mod 5)

X=1+5 t where t is any integer.

1+5 t=2 (mod 5)

5 t=1 (mod 5 )

5 is the Modular Inverse of modulo 5

t=1 (mod 5)

t =1+5 u : u= any integer.

X becomes 1+ 5 (1+5u )=6+25 u

6+25u =3 (mod 5)

25 u= -3 (mod 5)=2 (mod 5)

25 is Modular Inverse of Modulo 5

u=2 (mod 5)

u =2+5 v : v= any integer.

X becomes 6 +25 (2+5 v)=56+125 v .

56 +125 v=4 (mod 5)

125 v=-52 (mod 5)=3 (mod 5)

125 is Modular Inverse of modulo 5

v = 3 (mod 5)

v =3 +5 w where w is any integer.

For lowest 4 digit number, take

w=200

X=1003 : Lowest 4 digit number.

Take w=201

X=208 : Next number.

For 4 digit largest number, take

w=1995

X= 9978 : Largest number.

No. of Numbers =(1995 - 200 )+1=1800 □Ans.

8 0
3 years ago
Read 2 more answers
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