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Temka [501]
3 years ago
13

Evaluate the expression for x = -3 and y = 4 3x2 – 2xy

Mathematics
1 answer:
Lana71 [14]3 years ago
7 0

Answer: -36 is your answer

Step-by-step explanation:

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The scores of eight grade students in a math test are normally distributed with a mean of 57.5 and a standard deviation of 6.5.
rewona [7]
Work out the z scores  for   51  and 54 
z1 = (51 - 57.5) / 6.5 = -1
z2 = (64  - 57.5)/6.5 = 1
from tables of normal distribution this value os 3413 + 3413  = 68.26%


so the answer is  c
8 0
3 years ago
Joy is training for a triathlon. She runs a distance of 2x miles, cycles a distance of (x2 + 7) miles, and swims a distance of (
Tpy6a [65]
Add

2x+x²+7+x-2=
x²+2x+x+7-2=
x²+3x+5=total distance
6 0
3 years ago
Read 2 more answers
16x^2-25 and 4x^2-x-5?
Thepotemich [5.8K]
What are you asking?
6 0
3 years ago
What is the average rate of change of the function over the interval x = 0 to x = 8?
salantis [7]
Ans:  The average rate of change of function = \frac{1}{102} (in fraction)

Explanation:
The average rate of change of function  = \frac{f(b) - f(a)}{b - a}

Where
b = 8 (x's upper bound)
a = 0 (x's lower bound)

f(b) = f(8) =  \frac{2*8-2}{5*8-6} =  \frac{7}{17} \\ f(a) = f(0) =  \frac{2*0-2}{5*0-6} =  \frac{1}{3} \\  \frac{f(b)  - f(a)}{b-a} =  \frac{ \frac{7}{17} -  \frac{1}{3}}{8} =  \frac{1}{102}

Hence the average rate of change of function = \frac{1}{102} (in fraction)
5 0
3 years ago
A commuter crosses one of three bridges, A, B, or C, to go home from work. The commuter crosses bridge A with probability 1/3, c
mihalych1998 [28]

Answer:

0.1333 = 13.33% probability that bridge B was used.

Step-by-step explanation:

Conditional Probability

We use the conditional probability formula to solve this question. It is

P(B|A) = \frac{P(A \cap B)}{P(A)}

In which

P(B|A) is the probability of event B happening, given that A happened.

P(A \cap B) is the probability of both A and B happening.

P(A) is the probability of A happening.

In this question:

Event A: Arrives home by 6 pm

Event B: Bridge B used.

Probability of arriving home by 6 pm:

75% of 1/3(Bridge A)

60% of 1/6(Bridge B)

80% of 1/2(Bridge C)

So

P(A) = 0.75*\frac{1}{3} + 0.6*\frac{1}{6} + 0.8*\frac{1}{2} = 0.75

Probability of arriving home by 6 pm using Bridge B:

60% of 1/6. So

P(A \cap B) = 0.6*\frac{1}{6} = 0.1

Find the probability that bridge B was used.

P(B|A) = \frac{P(A \cap B)}{P(A)} = \frac{0.1}{0.75} = 0.1333

0.1333 = 13.33% probability that bridge B was used.

8 0
3 years ago
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