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trapecia [35]
3 years ago
9

(8x 2 −15x)−(x 2 −27x)=ax 2 +bxleft parenthesis, 8, x, squared, minus, 15, x, right parenthesis, minus, left parenthesis, x, squ

ared, minus, 27, x, right parenthesis, equals, a, x, squared, plus, b, x If the equation above is true for all values of xxx, what is the value of b-ab−ab, minus, a ?
Mathematics
1 answer:
quester [9]3 years ago
3 0

Answer:

<h2>5</h2>

Step-by-step explanation:

Given the expression (8x² −15x)−(x² −27x) = ax² +bx, we are to determine the value of b-a. Before we determine the vwlue of b-a, we need to first calculate for the value of a and b from the given expression.

On expanding the left hand side of the expression we have;

= (8x² −15x)−(x² −27x)

Open the paranthesis

= 8x² −15x−x²+27x

collect the like terms

= 8x²−x²+27x −15x

=  7x²+12x

Comparing the resulting expression with ax²+bx

7x²+12x =  ax²+bx

7x² = ax²

a = 7

Also;

12x = bx

b =12

The value of b - a = 12 - 7

b -a = 5

Hence the value of b-a is equivalent to 5

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devlian [24]

Answer:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

Step-by-step explanation:

Given the mean = 205 cm and standard deviation as 7.8cm

a. To calculate the probability that an individual distance is greater than 218.4 cm, we subtract the probability of the distance given (i.e 218.4 cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) from 1. Therefore, we have 1- P(Z\leq 1.72). Using the Z distribution table we have 1-0.9573. Therefore P(X >218.4)= 0.0427.

b. To calculate the probability that mean of 15 (i.e n=15) randomly selected distances is greater than 202.8, we subtract the probability of the distance given (i.e 202.8cm) from the mean (i.e 205 cm) divided by the standard deviation (i.e 7.8cm) divided by the square root of mean (i.e n= 15)  from 1. Therefore, we have 1- P(Z\leq -1.09). Using the Z distribution table we have 1-0.1378. Therefore P(X >202.8)= 0.8622.

c. This will also apply to a normally distributed data even if it is not up to the sample size of 30 since the sample distribution is not a skewed one.

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MissTica

Answer

B) When the x-value is 0, and the y-value is 1

Step-by-step explanation:

Edge

3 0
3 years ago
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