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kondor19780726 [428]
3 years ago
8

141÷6 (a) Write a story problem that matches this expression. (b) Solve the problem. (c) Does the quotient in your problem have

a remainder? If it does, explain what you did wirh the remainder when you solved the problem?
Mathematics
1 answer:
avanturin [10]3 years ago
8 0

There is a type of ammo in f/o/r/t/n/i/t/e that can be magically used for every single gun. You open an ammo box, and out pops 141 rounds of that ammo. You have already collected 6 guns. How many rounds go into each gun?

The answer is 23.5

THERE IS A REMAINDER??

Nothing happens! Literally! It never happened! There is no ammo that can be used for all the guns. GG and hope to see you on the game sometime, cause...i am not allowed to have it.

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You can tell that a fraction is a unit fraction when it is over a unit of one. This is also known as unit rate.
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Find the distance between (-18, 9) and (22,0).
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Answer:

Step-by-step explanation:

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An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an
igomit [66]

Answer:

Mean = 1.57

Variance=0.31

Step-by-step explanation:

To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:

1) Both surgeries are successful (X=2).

P(e_1)=0.90*0.67=0.603

2) Left knee unsuccessful and right knee successful (X=1).

P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067

3) Right knee unsuccessful and left knee successful (X=1).

P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297

4) Both surgeries are unsuccessful (X=0).

P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033

Then, the mean can be calculated as the expected value:

M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57

The variance can be calculated as:

V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111

3 0
3 years ago
Given: the function defined by f(x)=3x^2-4 Which statement is true? 1. f(0) = 0 2. f(-2) = f(2) 3. f(5)+ f(2) = f(7) 4. f(5) x f
sergeinik [125]
f(x)=3x^2-4 \\ \Downarrow \\ f(0)=3 \times 0^2-4=0-4=-4 \\ -4 \not= 0 \\ f(0) \not= 0 \\
\hbox{statement 1 is false} \\ \\
f(-2)=3 \times (-2)^2-4 =3 \times 4-4=12-4=8 \\
f(2)=3 \times 2^2-4=3 \times 4-4=12-4=8 \\ 8=8
\\ f(-2)=f(2) \\
\hbox{statement 2 is true}

f(5)=3 \times 5^2-4=3 \times 25-4=75-4=71 \\ f(2)=8 \\
f(7)=3 \times 7^2-4=3 \times 49-4=147-4=143 \\
71+8 \not= 143 \\
f(5)+f(2) \not= f(7) \\ \hbox{statement 3 is false} \\ \\
f(5)=71 \\ f(2)=8 \\ f(10)=3 \times 10^2-4=3 \times 100-4=300-4=296 \\
71 \times 8 \not= 296 \\
f(5) \times f(2) \not= f(10) \\
\hbox{statement 4 is false}

Stamement 2. f(-2)=f(2) is true.
3 0
3 years ago
PLEASE HELP WILL GIVE 10 POINTS. HURRYY
Anvisha [2.4K]

B. -4-1. -4-1=-5, which is the amount of you count the spaces in between.

Hope this helps!

4 0
3 years ago
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