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vovikov84 [41]
3 years ago
6

What is the solutions to 0=x^2-x-6 two things ,,quadratic equationz,,,

Mathematics
1 answer:
natita [175]3 years ago
3 0

Answer:

x1=3. x2=-2

Step-by-step explanation:

x =  \frac{ - b +  -  \sqrt{b {}^{2} - 4ac } }{2a}

b=-1

a=1

c=-6

x =  \frac{ - ( - 1) +  -  \sqrt{ {( - 1)}^{2} - 4 \times 1 \times ( - 6) } }{2 \times 1}

x =  \frac{1 +  -  \sqrt{1 + 24} }{2}

x =  \frac{1 + -   \sqrt{25} }{2}

x1 =  \frac{1 + 5}{2}

x2 =  \frac{1 - 5}{2}

x1=3

x2=-2

x =  \frac{1 +  - 5}{2}

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Write a system of equations to describe the situation below, solve using elimination, and fill in the blanks.
MissTica

Answer:

Blankets sell for $29 a piece, and hooded towels sell for $23 a piece.

Step-by-step explanation:

blanket = x

towel = y

5x + 13y = 444 -> x=(444-13y)/5

23x + 13y = 966

23[(443-13y)/5] +13y = 966

x = 29

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7 0
3 years ago
What is the solution of |5x| = 70?
kondaur [170]

Answer:

x=-14\:or\:x=14

Step-by-step explanation:

The given equation is

|5x|=70

This implies that;

5x=70\:or\:-5x=70

Solve for x to obtain;

x=\frac{70}{5}\:or\:x=\frac{70}{-5}

x=14\:or\:x=-14

7 0
3 years ago
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Aleksandr [31]
I think the answer is B
8 0
2 years ago
Suppose twenty-two communities have an average of = 123.6 reported cases of larceny per year. assume that σ is known to be 36.8
Delvig [45]
We are given the following data:

Average = m = 123.6
Population standard deviation = σ= psd = 36.8
Sample Size = n = 22

We are to find the confidence intervals for 90%, 95% and 98% confidence level.

Since the population standard deviation is known, and sample size is not too small, we can use standard normal distribution to find the confidence intervals.

Part 1) 90% Confidence Interval
z value for 90% confidence interval = 1.645

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.645* \frac{36.8}{ \sqrt{22}}=110.69

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.645* \frac{36.8}{ \sqrt{22}}=136.51

Thus the 90% confidence interval will be (110.69, 136.51)

Part 2) 95% Confidence Interval
z value for 95% confidence interval = 1.96

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-1.96* \frac{36.8}{ \sqrt{22}}=108.22

Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+1.96* \frac{36.8}{ \sqrt{22}}=138.98

Thus the 95% confidence interval will be (108.22, 138.98)

Part 3) 98% Confidence Interval
z value for 98% confidence interval = 2.327

Lower end of confidence interval = m-z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Lower end of confidence interval=123.6-2.327* \frac{36.8}{ \sqrt{22}}=105.34
Upper end of confidence interval = m+z *\frac{psd}{ \sqrt{n} }
Using the values, we get:
Upper end of confidence interval=123.6+2.327* \frac{36.8}{ \sqrt{22}}=141.86

Thus the 98% confidence interval will be (105.34, 141.86)


Part 4) Comparison of Confidence Intervals
The 90% confidence interval is: (110.69, 136.51)
The 95% confidence interval is: (108.22, 138.98)
The 98% confidence interval is: (105.34, 141.86)

As the level of confidence is increasing, the width of confidence interval is also increasing. So we can conclude that increasing the confidence level increases the width of confidence intervals.
3 0
3 years ago
Select the correct answer.
White raven [17]

Answer:

c is the best answer of this question

6 0
3 years ago
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