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3 years ago

Solve the equation for x, where x is a real number (5 points): 2x^2 + 7x - 9 = -3

1 answer:

6 0

Add 3 to both sides so that the equation becomes 2x^2 + 7x - 6 = 0.
To find the solutions to this equation, we can apply the quadratic formula. This quadratic formula solves equations of the form ax^2 + bx + c = 0
                              x = [ -b ± √(b^2 - 4ac) ] / (2a)
                              x = [ -7 ± √(7^2 - 4(2)(-6)) ] / ( 2(2) )
                              x = [ -7 ± √(49 - (-48) ) ] / ( 4 )
                              x = [ -7 ± √(97) ] / ( 4)
                              x = [ -7 ± sqrt(97) ] / ( 4 )
                              x = -7/4 ± sqrt(97)/4
The answers are -7/4 + sqrt(97)/4 and -7/4 - sqrt(97)/4.

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

X²=14x-45 step by step please

dolphi86 [110]

Answer:

X=9 and X=5

Step-by-step explanation:

So you want to get one side to equal zero so you can factor and find x. I recommend subtracting 14x and adding 45 to both sides so you don't have to deal with a negative quadratic.

X^2 - 14X +45 = 0

Now you can factor. Your looking for two factors that equal 45 and add to -14

All factors of 45:

1 and 45, 3 and 15, 5 and 9, -1 and -45,

-3 and -15, -5 and -9

So out of those combinations -5 and -9 both multiply to 45 and add up to -14 so these are our common factors

(X-5)(X-9)=0

X-5=0 add 5 to both sides X=5

X-9=0 add 9 to both sides X=9

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3 years ago

Which expression represents how many points each worksheet is worth?

photoshop1234 [79]

Answer:

105 divided by 7

Step-by-step explanation:

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3 years ago

Read 2 more answers

The amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and standard deviation 13 mL. Supp

andreyandreev [35.5K]

Answer:

(a) X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

$\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

Step-by-step explanation:

We are given that the amount of syrup that people put on their pancakes is normally distributed with mean 63 mL and a standard deviation of 13 mL.

Suppose that 43 randomly selected people are observed pouring syrup on their pancakes.

(a) Let X = amount of syrup that people put on their pancakes

The z-score probability distribution for the normal distribution is given by;

Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

So, the distribution of X ~ N( $\mu=63, \sigma^{2} = 13^{2}$ ).

Let $\bar X$ = sample mean amount of syrup that people put on their pancakes

The z-score probability distribution for the sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = mean amount of syrup = 63 mL

$\sigma$ = standard deviation = 13 mL

n = sample of people = 43

So, the distribution of $\bar X$ ~ N( $\mu=63,s^{2} = (\frac{13}{\sqrt{43} } )^{2}$ ).

(b) If a single randomly selected individual is observed, the probability that this person consumes is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < X < 62.8 mL)

P(61.4 mL < X < 62.8 mL) = P(X < 62.8 mL) - P(X $\leq$ 61.4 mL)

P(X < 62.8 mL) = P( $\frac{X-\mu}{\sigma}$ < $\frac{62.8-63}{13}$ ) = P(Z < -0.02) = 1 - P(Z $\leq$ 0.02)

= 1 - 0.50798 = 0.49202

P(X $\leq$ 61.4 mL) = P( $\frac{X-\mu}{\sigma}$ $\leq$ $\frac{61.4-63}{13}$ ) = P(Z $\leq$ -0.12) = 1 - P(Z < 0.12)

= 1 - 0.54776 = 0.45224

Therefore, P(61.4 mL < X < 62.8 mL) = 0.49202 - 0.45224 = 0.0398.

(c) For the group of 43 pancake eaters, the probability that the average amount of syrup is between 61.4 mL and 62.8 mL is given by = P(61.4 mL < $\bar X$ < 62.8 mL)

P(61.4 mL < $\bar X$ < 62.8 mL) = P( $\bar X$ < 62.8 mL) - P( $\bar X$ $\leq$ 61.4 mL)

P( $\bar X$ < 62.8 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{62.8-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z < -0.10) = 1 - P(Z $\leq$ 0.10)

= 1 - 0.53983 = 0.46017

P( $\bar X$ $\leq$ 61.4 mL) = P( $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{61.4-63}{\frac{13}{\sqrt{43} } }$ ) = P(Z $\leq$ -0.81) = 1 - P(Z < 0.81)

= 1 - 0.79103 = 0.20897

Therefore, P(61.4 mL < X < 62.8 mL) = 0.46017 - 0.20897 = 0.2512.

(d) Yes, for part (d), the assumption that the distribution is normally distributed necessary.

4 0

3 years ago