Ask question

Ask question

All categories

2 years ago

12

What is the range for the following set of measurements? 7.1 g, 9.8 g, 2.3 g, 8.5 g, 7.4 g, 5.7 g

1 answer:

tankabanditka [31]2 years ago

8 0

Answer:

7.5

Step-by-step explanation:

you take the largest number (9.8) and subtract with the smallest number (2.3)

<h2>9.8-2.3= 7.5</h2>

You might be interested in

The graph of which function passes through (0,3) and has an amplitude of 3?

evablogger [386]

The choices are supposed to be
f(x) = sin x + 3
f(x) = cos x + 3
f(x) = 3 sin x
f(x) = 3 cos x

The amplitude is the value of the numerical coefficient of sin or cos. The only possible answers are
f(x) = 3 sin x
f(x) = 3 cos x

Next, the function must pass through the point (0,3)
3 sin 0 = 0 and
3 cos = 3

Therefore, the answer is
f(x) = 3 cos x<span />

6 0

3 years ago

Read 2 more answers

F(x) = 2x-(x+6) for f(-3)

ICE Princess25 [194]

Plug in -3 for x
f(-3) = 2(-3) - (-3 + 6)
f(-3) = -6 - (3)
f(-3) = -9
Solution: f(-3) = -9

4 0

3 years ago

Read 2 more answers

Find the smallest of two consecutive even integers if the square is 10 more than the larger

djverab [1.8K]

I think your answers are 4 and 6

7 0

3 years ago

A computer repairman charges $30 to come to a home or office , plus $22 per hour of work.During one week, he visits 15 homes or

timofeeve [1]

31 hours.
first I times $30 by 15, an I got 450, then I got 1132 and took away 450 and I got 682, finally I divide 682 by 22 and got 31

3 0

3 years ago

Average box of crackers is 24.5 ounces with standard deviation of. 8 ounce. What percent of the boxes weigh more than 22.9 ounce

34kurt

Answer:

97.7% of of the boxes weigh more than 22.9 ounces.

15.9% of of the boxes weigh less than 23.7 ounces.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 24.5 ounces

Standard Deviation, σ = 0.8 ounce

We are given that the distribution of boxes weight is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

a) P(boxes weigh more than 22.9 ounces)

P(x > 22.9)

$P( x > 22.9) = P( z > \displaystyle\frac{22.9 - 24.5}{0.8}) = P(z > -2)$

$= 1 - P(z \leq -2)$

Calculation the value from standard normal z table, we have,

$P(x > 22.9) = 1 - 0.023 =0.977= 97.7\%$

97.7% of of the boxes weigh more than 22.9 ounces.

b) P(boxes weigh less than 23.7 ounces)

P(x < 23.7)

$P( x < 23.7) = P( z < \displaystyle\frac{23.7 - 24.5}{0.8}) = P(z < -1)$

Calculation the value from standard normal z table, we have,

$P(x < 23.7) =0.159= 15.9\%$

15.9% of of the boxes weigh less than 23.7 ounces.

7 0

3 years ago