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2 years ago

Which statement is true about the graphed function?

Mathematics

2 answers:

Ksenya-84 [330]2 years ago

7 0

Answer:

Step-by-step explanation:

Nady [450]2 years ago

3 0

Answer:

the last option or D

Step-by-step explanation:

f(x) > 0 over the interval (-∞, -4)

i hope this helps! quizlet is also very helpful lol, have a wonderful rest of your day/night, xx, nm <3 :)

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Using the quadratic formula to solve 7x2 -x = 7, what are the values of x?

Sunny_sXe [5.5K]

Answer:

x=-7

Step-by-step explanation:

7x2-x=7

7x2=14.

14-x=7

When you finished the last step, you will subtract 14 to both sides.

14-x=7

14-14-x=7-14

this will cancel out the 14 on the left.

And will leave you with:

-x=7-14

7-14 is -7.

and this will get you to

-x=-7

To finish this off, you have to turn the negative x/variable, into a positive x/variable.

So, you will divide both sides of the equation by the same term.

-x = -7

into

$\frac{-x}{1} =\frac{-7}{1}$

and it will be -7.

I hope this helps :)

8 0

3 years ago

The repeating decimal number 8.2151515... written as a fraction is

Blababa [14]

Closest would be 8 43/200

4 0

3 years ago

How do you add add fraction with a different denominator?

kenny6666 [7]

Find the LCM (least common multiple) of the denominators, and use that as the denominator for the two fractions.

Ex.

$\frac{3}{4} +\frac{2}{3}$

LCM (3, 4) is 12

$\frac{9}{12} +\frac{8}{12}$

= $\frac{17}{12}$

3 0

2 years ago

Read 2 more answers

HELPPPPP MEEEEEE Which one is not a solution to y = 3x - 5? A. (0, –5) B. (– 1, – 2) C. (5/3, 0) D. (2, 1)

nignag [31]

Answer:

Step-by-step explanation:

Let's examine the individual things

A. x=0 --> y=3.0-5=-5 True

B. x=-1 --> y=3.(-1)-5=-8 False

C. x=5/3 -->y=3(5/3)-5=0 True

D. x=2 --> y=3.2-5=1 True

Answer is B.

You don't understand, you can ask me :)

7 0

3 years ago

A simple random sample of size n is drawn from a population that is normally distributed. The sample mean, x overbar, is found

joja [24]

Answer:

(a) 80% confidence interval for the population mean is [109.24 , 116.76].

(b) 80% confidence interval for the population mean is [109.86 , 116.14].

(c) 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed.

Step-by-step explanation:

We are given that a simple random sample of size n is drawn from a population that is normally distributed.

The sample mean is found to be 113 and the sample standard deviation is found to be 10.

(a) The sample size given is n = 13.

Firstly, the pivotal quantity for 80% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean = 113

s = sample standard deviation = 10

n = sample size = 13

$\mu$ = population mean

Here for constructing 80% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 80% confidence interval for the population mean, $\mu$ is ;

P(-1.356 < $t_1_2$ < 1.356) = 0.80 {As the critical value of t at 12 degree

of freedom are -1.356 & 1.356 with P = 10%}

P(-1.356 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.356) = 0.80

P( $-1.356 \times }{\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

P( $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ) = 0.80

80% confidence interval for $\mu$ = [ $\bar X-1.356 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.356 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.356 \times }{\frac{10}{\sqrt{13} } }$ , $113+1.356 \times }{\frac{10}{\sqrt{13} } }$ ]

= [109.24 , 116.76]

Therefore, 80% confidence interval for the population mean is [109.24 , 116.76].

(b) Now, the sample size has been changed to 18, i.e; n = 18.

So, the critical values of t at 17 degree of freedom would now be -1.333 & 1.333 with P = 10%.

80% confidence interval for $\mu$ = [ $\bar X-1.333 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+1.333 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-1.333 \times }{\frac{10}{\sqrt{18} } }$ , $113+1.333 \times }{\frac{10}{\sqrt{18} } }$ ]

= [109.86 , 116.14]

Therefore, 80% confidence interval for the population mean is [109.86 , 116.14].

(c) Now, we have to construct 98% confidence interval with sample size, n = 13.

So, the critical values of t at 12 degree of freedom would now be -2.681 & 2.681 with P = 1%.

98% confidence interval for $\mu$ = [ $\bar X-2.681 \times }{\frac{s}{\sqrt{n} } }$ , $\bar X+2.681 \times }{\frac{s}{\sqrt{n} } }$ ]

= [ $113-2.681 \times }{\frac{10}{\sqrt{13} } }$ , $113+2.681 \times }{\frac{10}{\sqrt{13} } }$ ]

= [105.56 , 120.44]

Therefore, 98% confidence interval for the population mean is [105.56 , 120.44].

(d) No, we could not have computed the confidence intervals in parts (a)-(c) if the population had not been normally distributed because t test statistics is used only when the data follows normal distribution.

6 0

3 years ago