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3 years ago

Solve the equation

Mathematics

1 answer:

kondor19780726 [428]3 years ago

4 0

4 1/5 + B = 9 3/5

21/5 + B = 48/5

B = 27/5

B = 5 2/5

Letter C is the correct answer!

Hope this helps! ;)

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A sample of 81 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of th

Kruka [31]

Answer:

95% confidence interval for the true mean score is [4.6 , 6.6].

Step-by-step explanation:

We are given that a sample of 81 tobacco smokers who recently completed a new smoking-cessation program were asked to rate the effectiveness of the program on a scale of 1 to 10.

The average rating was 5.6 and the standard deviation was 4.6.

Firstly, the pivotal quantity for 95% confidence interval for the true mean is given by;

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

where, $\bar X$ = sample average rating = 5.6

s = sample standard deviation = 4.6

n = sample of tobacco smokers = 81

$\mu$ = population mean score

Here for constructing 95% confidence interval we have used One-sample t test statistics as we don't know about population standard deviation.

So, 95% confidence interval for the population mean score, $\mu$ is ;

P(-1.993 < $t_8_0$ < 1.993) = 0.95 {As the critical value of t at 80 degree of

freedom are -1.993 & 1.993 with P = 2.5%}

P(-1.993 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.993) = 0.95

P( $-1.993 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.993 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.993 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.993 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.993 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.993 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $5.6-1.993 \times {\frac{4.6}{\sqrt{81} } }$ , $5.6+1.993 \times {\frac{4.6}{\sqrt{81} } }$ ]

= [4.6 , 6.6]

Therefore, 95% confidence interval for the true mean score is [4.6 , 6.6].

4 0

2 years ago

Help me plz I need help

HACTEHA [7]

Answer:

5 5/8

Step-by-step explanation:

3 0

2 years ago

Read 2 more answers

One number is 2 less than a second number. twice the second number is 16 more than 4 times the first. find the two numbers.

Nostrana [21]

Let the numbers be x and y

x= y-2.... (1)
and, 2y = 16 + 4x.... (2)

putting the value of x from (1) in (2),

2y = 16 + 4x
=> 2y = 16 + 4(y-2)
=> 2y = 16 + 4y - 8
=> 2y = 8 + 4y
=> 2y - 4y = 8
=> -2y = 8
=> y =-8/2
=> y= -4

and x = -4-2 = -6

hence , in short the numbers are -2 and -4

4 0

2 years ago

Please help me on this , will give brainliest , tyy

sweet-ann [11.9K]

Answer:

Step-by-step explanation:

7 0

2 years ago

Solve the following System Of Equations Using Substitution or Elimination Methods. Show Work.

Ostrovityanka [42]

Answer:

There are two pairs of solutions: (2,7) and (-1,4)

Step-by-step explanation:

We will use substitution.

y = x^2 + 3

y = x +5

Since the second equation is equal to y, replace y in the first equation with the second equation.

y = x^2 + 3

x + 5 = x^2 + 3

Rearrange so that one side is equal to 0.

5 - 3 = x^2 - x

2 = x^2 - x

0 = x^2 - x - 2

You may use quadratic formula or any form of factoring to find the zeros (x values that make the equation equal to 0).

a = 1, b = -1, c = -2

Zeros = $\frac{-b + \sqrt{b^{2}-4ac } }{2a}$ and $\frac{-b - \sqrt{b^{2}-4ac } }{2a}$

Zeros = 2 and -1

Now that you have your x values, plug them into the equations to find their corresponding y values.

y = x^2 + 3

y = (2)^2 + 3

y = 7

Pair #1: (2,7)

y = x^2 + 3

y = (-1)^2 + 3

y = 4

Pair #2: (-1,4)

Therefore, there are two pairs of solutions: (2,7) and (-1,4).

8 0

2 years ago