Which of these situations can be represented by the opposite of

Please help

ololo11 [35]

Answer:

8

Step-by-step explanation:

3 + 5 = 8

3 0

3 years ago

Use the equation to complete an algebraic proof that proves the answer is x = 7/6. Write your proof in your journal and upload y

Komok [63]

Answer:

x=7/6

Step-by-step explanation:

subtract the 4x from both sides of the equation

(2x+6)-4x=4x-3-4x

find common denominator

(2x+6)/5+(5(-4)x)/5=4x-3-4x

then combine fractions

(2x+6+5(-4)x)/5=4x-3-4x

then multiply the 5&4

(2x+6-20x)/5= 4x3-4x

combine terms

(-18x+6)/5= -3

then multiply everything by 5

-18x+6= -15

subtract the 6 from both sides

-18x= -21

divide -21 by -18

(-21)/(-18)=x

find greatest common multiple (3) and solve

x = 7/6

5 0

3 years ago

The width of a vegetable garden is į times its length.

e-lub [12.9K]

Answer:

I guess it's 6 but I'm not really sure about my answer

8 0

2 years ago

The heights of 40 randomly chosen men are measured and found to follow a normal distribution. An average height of 175 cm is obt

AVprozaik [17]

Answer:

95% two-sided confidence interval for the true mean heights of men is [168.8 cm , 181.2 cm].

Step-by-step explanation:

We are given that the heights of 40 randomly chosen men are measured and found to follow a normal distribution.

An average height of 175 cm is obtained. The standard deviation of men's heights is 20 cm.

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

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

where, $\bar X$ = sample average height = 175 cm

$\sigma$ = population standard deviation = 20 cm

n = sample of men = 40

Here for constructing 95% confidence interval we have used One-sample z test statistics as we know about population standard deviation.

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

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ , $\bar X+1.96 \times }{\frac{\sigma}{\sqrt{n} } }$ ]

= [ $175-1.96 \times }{\frac{20}{\sqrt{40} } }$ , $175+1.96 \times }{\frac{20}{\sqrt{40} } }$ ]

= [168.8 cm , 181.2 cm]

Therefore, 95% confidence interval for the true mean height of men is [168.8 cm , 181.2 cm].

The interpretation of the above interval is that we are 95% confident that the true mean height of men will be between 168.8 cm and 181.2 cm.

3 0

3 years ago

Which statement is not true about the data shown by the box plot below?

hodyreva [135]

Answer:

B. The interquartile range is 55

Step-by-step explanation:

the interquartile range is Q3 minus Q1, you would get 15 ,not 55. So B is the answer to your question

7 0

2 years ago

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Which of these situations can be represented by the opposite of - 25