Is 3/15 repeating decimal

Fifteen hungry kittens found a bucket of sausages. Each time

raketka [301]

Answer:

2.4 SAUSAGES

Step-by-step explanation:

18 x 2 = 36

36/ 15 = 2.4

6 0

1 year ago

SOMEONE PLZ HELP SO I CAN NOT GET A 10% again :) thanks I will let u hit my cart

Step2247 [10]

Answer:

it's "c" or the third option

4 0

2 years ago

How to find average value of a function over a given interval?

Strike441 [17]

f(x)=8x−6f(x)=8x-6 , [0,3][0,3]

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.(−∞,∞)(-∞,∞){x|x∈R}{x|x∈ℝ}f(x)f(x) is continuous on [0,3][0,3].f(x)f(x) is continuousThe average value of function ff over the interval [a,b][a,b] is defined as A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dx.A(x)=1b−a∫baf(x)dxA(x)=1b-a∫abf(x)dxSubstitute the actual values into the formula for the average value of a function.A(x)=13−0(∫308x−6dx)A(x)=13-0(∫038x-6dx)Since integration is linear, the integral of 8x−68x-6 with respect to xx is ∫308xdx+∫30−6dx∫038xdx+∫03-6dx.A(x)=13−0(∫308xdx+∫30−6dx)A(x)=13-0(∫038xdx+∫03-6dx)Since 88 is constant with respect to xx, the integral of 8x8x with respect to xx is 8∫30xdx8∫03xdx.A(x)=13−0(8∫30xdx+∫30−6dx)A(x)=13-0(8∫03xdx+∫03-6dx)By the Power Rule, the integral of xx with respect to xx is 12x212x2.A(x)=13−0(8(12x2]30)+∫30−6dx)

3 0

3 years ago

The histogram below shows the distribution of the assets (in millions of dollars) of 71 companies. Does the distribution appear

solniwko [45]

Answer:

Yes, the assets appear to follow a normal distribution, the values are concentrated in the center and taper off towards the ends

Step-by-step explanation:

The distribution shown above is normal as it exhibits symmetry. This means thatvtge values are concentrated in the middle with the peak so situated in the middle of the distribution which is exactly what is displayed above. As we move towards either side of the center, the values begin to decrease and we have the tail at either side of the midpoint and not on one side of the distribution.

7 0

2 years ago

For how many real values of x is <img src="https://tex.z-dn.net/?f=%20%5Csqrt%7B120-%5Csqrt%7Bx%7D%7D%20" id="TexFormula1" title

leva [86]

A good place to start is to set $\sqrt{x}$ to y. That would mean we are looking for $\sqrt{120-y}$ to be an integer. Clearly, $y\leq 120$ , because if y were greater the part under the radical would be a negative, making the radical an imaginary number, not an integer. Also note that since $\sqrt{x}$ is a radical, it only outputs values from $[0,\infty]$ , which means y is on the closed interval: $[0,120]$ .

With that, we don't really have to consider y anymore, since we know the interval that $\sqrt{x}$ is on.

Now, we don't even have to find the x values. Note that only 11 perfect squares lie on the interval $[0,120]$ , which means there are at most 11 numbers that x can be which make the radical an integer. All of the perfect squares are easily constructed. We can say that if k is an arbitrary integer between 0 and 11 then:

$\sqrt{120-\sqrt{x}}=k \implies \\ \sqrt{x}=k^2-120 \implies\\ x=(k^2-120)^2$

Which is strictly positive so we know for sure that all 11 numbers on the closed interval will yield a valid x that makes the radical an integer.

5 0

2 years ago