What do you call it when a bunch of kids throw circles at eachother

Nick found the quotient of 8.64 and 1.25 × 105. His work is shown below. 1. (8.64 × 101) (1.25 × 105) 2. (8.64 1.25 ) (101 105 )

Savatey [412]

Answer:

B.) No, the power multiplied to 8.64 should have an exponent of 0.

Step-by-step explanation:

took the test on Edgenuity and got it right! Hope this helps and please mark brainliest if can.

4 0

1 year ago

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The box which measures 70cm X 36cm X 12cm is to be covered by a canvas. How many meters of canvas of width 80cm would be require

grigory [225]

Answer:

142.2 meters.

Step-by-step explanation:

We have been given that a box measures 70 cm X 36 cm X 12 cm is to be covered by a canvas.

Let us find total surface area of box using surface area formula of cuboid.

$\text{Total surface area of cuboid}=2(lb+bh+hl)$ , where,

$l$ = Length of cuboid,

$b$ = Breadth of cuboid,

$w$ = Width of cuboid.

$\text{Total surface area of box}=2(70\cdot36+36\cdot 12+12\cdot 70)$

$\text{Total surface area of box}=2(2520+432+840)$

$\text{Total surface area of box}=2(3792)$

$\text{Total surface area of box}=7584$

Therefore, the total surface area of box will be 7584 square cm.

To find the length of canvas that will cover 150 boxes, we will divide total surface area of 150 such boxes by width of canvass as total surface area of canvas will also be the same.

$\text{Width of canvas* Length of canvass}=\text{Total surface area of 150 boxes}$

$80\text{ cm}\times\text{ Length of canvass}=150\times 7584\text{cm}^2$

$\text{ Length of canvass}=\frac{150\times 7584\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=\frac{1137600\text{ cm}^2}{80\text{ cm}}$

$\text{ Length of canvass}=14220\text{ cm}$

Let us convert the length of canvas into meters by dividing 14220 by 100 as 1 meter equals to 100 cm.

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100\frac{cm}{m}}$

$\text{ Length of canvass}=\frac{14220\text{ cm}}{100}\times\frac{m}{cm}$

$\text{ Length of canvass}=142.20\text{ m}$

Therefore, 142.2 meters of canvas of width 80 cm required to cover 150 such boxes.

5 0

2 years ago

Researchers have a sample size of 24 , and they are using the Student's t-distribution. What are the degrees of freedom and how

DerKrebs [107]

Answer:

For this case assuming that the random variable is X

$df = n-1$

And replacing n = 24 we got:

$df = 24-1=23$

And we notate the distribution we got: $X \sim t_{n-1}= t_{23}$

Step-by-step explanation:

Previous concepts

The t distribution (Student’s t-distribution) is a "probability distribution that is used to estimate population parameters when the sample size is small (n<30) or when the population variance is unknown".

The shape of the t distribution is determined by its degrees of freedom and when the degrees of freedom increase the t distirbution becomes a normal distribution approximately.

The degrees of freedom represent "the number of independent observations in a set of data. For example if we estimate a mean score from a single sample, the number of independent observations would be equal to the sample size minus one."

Solution to the problem

For this case assuming that the random variable is X

$df = n-1$

And replacing n = 24 we got:

$df = 24-1=23$

And we notate the distribution we got: $X \sim t_{n-1}= t_{23}$

7 0

1 year ago

Which statement is true about the discontinuities of the function f(x)?

Tema [17]

Answer:

There are asymptotes at x = three-halves and x = negative one-third.

Step-by-step explanation:

f(x) = (x + 1)/ (6x^2 - 7x - 3)

= (x + 1)( / (6x^2 + 2x - 9x - 3)

= (x + 1) / (2x(3x + 1) - 3(3x + 1))

= (x + 1) / (2x - 3)(3x + 1)

Now x = 3/2 and x = -1/3 both make te denominator zero so these are both asymptotes.

6 0

1 year ago

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Cameron wondered if the average score on a final exam was different between those who texted on a regular basis during the lectu

Margarita [4]

Answer:

c. A two-tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

Step-by-step explanation:

Let p1 be the average score on a final exam who texted on a regular basis during the lectures for a particular class

And p2 be the average score on a final exam who did not texted at all during the lectures for a particular class

According to the Cameron's point of interest, null and alternative hypotheses are:

$H_{0}:$ p1 = p2

$H_{a}:$ p1 ≠ p2

Two tailed test should be performed since the alternative hypothesis states that the parameter is not equal to the hypothesized value.

3 0

2 years ago