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Sliva [168]
3 years ago
10

Christy correctly answer 85% of the questions on her math test if she missed 6 problems how many questions were there on the tes

t?
Mathematics
1 answer:
natta225 [31]3 years ago
3 0
Let total number of questions be x.

If she answered 85% correctly, than that means she answered 15% incorrectly.

You can set up the equation 15% (0.15) of total number of questions (x) equals 6.
0.15*x = 6
x = 40

There were 40 questions on the test.
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Simplify the expression by combining like terms<br>4r+3c-2r+9c​
Novay_Z [31]

Answer:

2r+12c

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
What is the probability that e) A fair coin lands Heads 6 times in a row? f) A fair coin lands Heads 4 times out of 5 flips? g)
cricket20 [7]

Answer:

e) 1.56%

f) 15.62%

h) 0.879%

g) 11.72%

Step-by-step explanation:

What we will do is solve point by point.

e)  A fair coin lands Heads 6 times in a row?

We have the following:

Total number of possible outcomes = 2 ^ 6 = 64

Number of favorable outcomes = 1

Required probability = 1/64 = 1.56%

f) A fair coin lands Heads 4 times out of 5 flips

We have the following:

Total number of possible outcomes = 2 ^ 5 = 32

Number of favorable outcomes = 5C4

nCr = n! / (r! * (n-r)!)

5C4 = 5! / (4! * (5-4)!) = 5

Required probability = 5/32 = 15.62%

g) he bit string has exactly two 1s, given that the string begins with a 1 if you pick a bit string from the set of all bit strings of length ten?

We have the following:

Total number of possible outcomes = 2 ^ 10 = 1024

Number of ways in which a position excluding the start of the string can be chosen is 9C1

Total number of favorable outcomes = 9C1

9C1 = 9! / (1! * (9-1)!) = 9

Required probability = 9/1024  = 0.879%

h)The bit string has the sum of its digits equal to seven if you pick a bit string from the set of all bit strings of length ten?

We have the following:

Total number of possible outcomes = 2 ^ 10 = 1024

For the sum of the digits to be 7 there has to be 7 ones.

Number of ways in which 7 position can be chosen is 10C7.

Total number of favorable outcomes = 10C7

10C7 = 10! / (7! * (10-7)!) = 120

Required probability = 120/1024 = 11.72%

5 0
4 years ago
Many residents of suburban neighborhoods own more than one car but consider one of their cars to be the main family vehicle. The
horrorfan [7]

Answer:

95.4% of family vehicles is between 1 and 3 years old.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 2

Standard Deviation, σ = 6 months = 0.5 year

We are given that the distribution of age of cars is a bell shaped distribution that is a normal distribution.

Formula:

z_{score} = \displaystyle\frac{x-\mu}{\sigma}

P(family vehicles is between 1 and 3 years old)

P(1 \leq x \leq 3)\\\\ = P(\displaystyle\frac{1 - 2}{0.5} \leq z \leq \displaystyle\frac{3-2}{0.5}) = P(-2 \leq z \leq 2)\\\\= P(z \leq 2) - P(z < -2)\\= 0.977 -0.023 = 0.954= 95.4\%

P(1 \leq x \leq 3) = 95.4%

95.4% of family vehicles is between 1 and 3 years old.

3 0
3 years ago
What is the distance between the points (4,6) and (7,10)
arsen [322]

Answer:

(3,4)

Step-by-step explanation:

Did (7,10) - (4,6)

3 0
4 years ago
Discrete math
nadya68 [22]

Answer:

1) yes

2) No

3) No

4) yes

5) No

Step-by-step explanation:

1) f(0)=0, f(1)=1, f(2)=2, f(3)=3, then f[\{0, 1, 2, 3\}]=\{0, 1, 2, 3\}

2) f(0)=0, 1^2=1\equiv 1 \text{mod 4}, then f(1)=1; 2^2=4\equiv 0 \text{ mod 4}, then f(2)=0; 3^2=9\equiv 1 \text{mod 4}, then f(3)=1

Then f[\{0, 1, 2, 3\}]=\{0, 1, 3\}, this means that f isn't onto.

3.

  • f(0)=0;
  • 1^1-1=0, then f(1)=0
  • 2^2-2=2, then f(2)=2
  • 3^2-3=6\equiv 2 \text{ mod 4}, then f(3)=2

Then  f[\{0, 1, 2, 3\}]=\{0, 2\}, this means that f isn't onto.

4.  f[\{0, 1, 2, 3\}]=\{0, 1, 2,3\}, then f is onto.

5. f[\{0, 1, 2, 3\}]=\{1, 2\}, this means that f isn't onto.

6 0
3 years ago
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