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3 years ago

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Joyce is trying to solve the equation y = x2 − 8x + 7 using the quadratic formula. She has made an error in one of the steps bel

ow. Find the step where Joyce went wrong.

1 answer:

kobusy [5.1K]3 years ago

4 0

Answer:

x = 1 , 7

Step-by-step explanation:

Solution:-

- The given equation is as follows:

y = x^2 - 8x + 7

- We can solve the above equation by either making factors or by using Quadratic formula.

Factor Approach:

- Using the constant "7" at the end of the quadratic equation we will determine two integer multiples such that their additions/subtraction results in "-8".

- So the only factor of "7" are:

7 x 1 = 7

-7 x -1 = 7

- We see that addition/subtraction of first (7 , 1 ) does not results in "-8", However, the sum of ( -1 , -7 ) = -1 - 7 = -8. So the correct factors are ( -1 , -7 ). So we replace "-8x" with our factors "-1x" and "-7x":

x^2 -x -7x + 7 = 0

- Take common multiples out of pair of two terms:

x*(x-1) -7*(x-1) = 0

(x-7)*(x-1) = 0

- So we equate each term in bracket with "0" and evaluate the values of x:

(x-7) = 0 , x = 7

(x-1) = 0 , x = 1

- So the solution to the quadratic equation is:

x = 1 , 7

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A press release describing the 1986 nonstop, round-the-world trip by the aircraft voyager including the following data:

adoni [48]

Answer:

a.) the average speed of the aircraft in miles per hour

115.76 miles/ hour

b.) the fuel consumption in miles per pound of fuel

(assume a density of 0.70 g/mL for the fuel.)

2.8 miles/Ib of fuel

Step-by-step explanation:

A press release describing the 1986 nonstop, round-the-world trip by the aircraft voyager including the following data:

flight distance: 25,012 miles

flight time: 9 days, 3 min., 44 seconds

fuel capacity: nearly 9000 lbs.

fuel remaining at the end of flight: 14 gal.

To the maximum number of significant figure permitted, calculate:

a.) the average speed of the aircraft in miles per hour

Step 1

We convert the time to hours

flight time: 9 days, 3 min., 44 seconds

1 day = 24 hours

9 days =

24 × 9

= 216 hours

60 mins = 1 hour

3 mins =

3/60 = 1/20 = 0.05 hours

1 hour = 60 minutes

1 minute = 60 seconds

1 hour = 60 × 60 = 3600 seconds

If 3600 seconds = 1 hour

44 seconds =

= 44/3600

= 0.0122222222 hours

Total number of hours spent = (216 + 0.05 + 0.0122222222) hours

= 216.06222222 hours

Speed = Distance/Time

= flight distance: 25,012 miles

Speed = 25,012 miles/ 216.06222222hours

= 115.76294895 miles/ hour

Approximately = 115.76 miles/ hour

b.) the fuel consumption in miles per pound of fuel

(assume a density of 0.70 g/mL for the fuel.)

Step 1

We determine how much fuel is used

We have 14 gallons of fuel left over

1 gallon = 3.785 litres

14 gallons =

Cross Multiply

14 gallons × 3.785 liters/1 gallon

= 52.9958 liters

Note that density of fuel =

0.70 g/ml

1 gram / millilitre = 1 kilogram / litre

Hence, 0.7g/ml = 0.7 kg/L

Hence,the mass of fuel left = 52.9958 liters * 0.7 = 37.09706kg

1 kilogram = 2.205 pounds

37.09706kg =

37.09706kg × 2.205 pounds

= 81.7990173 lbs

Since the full capacity = 9000 lbs

The fuel used = 9000 - 81.7990173 lbs

= 8918.2009827 lbs

Hence, fuel consumption is

Distance/ amount of fuel used

25012/8918.2009827 lbs = 2.8046015164 miles/lb of fuel

Approximately = 2.8 miles/Ib of fuel

7 0

3 years ago

The amount of money spent on textbooks per year for students is approximately normal.

Ostrovityanka [42]

Answer:a

a

$336.04 < \mu < 443.96$

b

The margin of error will increase

c

The margin of error will decreases

d

The 99% confidence interval is $0.4107 < p < 0.4293$

Step-by-step explanation:

From the question we are told that

The sample size $n = 19$

The sample mean is $\= x = \$\ 390$

The standard deviation is $\sigma = \$ \ 120$

Given that the confidence level is 95% then the level of significance is mathematically represented as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table

So

$Z_{\frac{\alpha }{2} } = 1.96$

The margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

=> $E = 1.96 * \frac{120}{\sqrt{19} }$

=> $E = 53.96$

The 95% confidence interval is

$\= x - E < \mu < \= x + E$

=> $390 - 53.96 < \mu < 390 - 53.96$

=> $336.04 < \mu < 443.96$

When the confidence level increases the $Z_{\frac{\alpha }{2} }$ also increases which increases the margin of error hence the confidence level becomes wider

Generally the sample size mathematically varies with margin of error as follows

$n \ \ \alpha \ \ \frac{1}{E^2 }$

So if the sample size increases the margin of error decrease

The sample proportion is mathematically represented as

$\r p = \frac{210}{500}$

$\r p = 0.42$

Given that the confidence level is 0.99 the level of significance is $\alpha = 0.01$

The critical value of $\frac{\alpha }{2}$ from the normal distribution table is

$Z_{\frac{\alpha }{2} } = 2.58$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} }* \sqrt{ \frac{\r p (1- \r p )}{n} }$

=> $E = 0.42 * \sqrt{ \frac{0.42 (1- 0.42 )}{ 500} }$

=> $E = 0.0093$

The 99% confidence interval is

$\r p - E < p < \r p + E$

$0.42 - 0.0093 < p < 0.42 + 0.0093$

$0.4107 < p < 0.4293$

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Answer:

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Step-by-step explanation:

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