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4 years ago

The probability of spinning an even number is 40%. What is the probability of not spinning an even number written as a decimal?

Mathematics

1 answer:

Lorico [155]4 years ago

3 0

That would be .6 because that means 60 percent.

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Is 2 an irrational number

Lena [83]

Answer:

no its rational

Step-by-step explanation:

3 0

3 years ago

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Rewrite the following polynomial in standard form

blagie [28]

Answer:

-7x³ - ⅑x - 1

Step-by-step explanation:

A polynomial written in standard form would be written in such a way that the term with the highest degree comes first, followed by the second highest and so on, then what comes last is the constant.

Given the polynomial -7x³ - 1 - ⅑x, -7x³ with the highest degree comes first, followed by -⅑x, and then the constant, -1, comes last.

Thus, we would have:

✅-7x³ - ⅑x - 1

8 0

3 years ago

Hattie accepted a job as a makeup artist after being offered a $550 signing

Radda [10]

The equation would be y=32x+550. 32 an hour is represented by 32x since she earned 32 an hour. 550 is the extra amount she is getting paid.

I hope this helps.

7 0

3 years ago

100 POINTS FOR BRAINLIEST! according to the image, whats the answer? Answer wisely for the full 100

lesantik [10]

Answer:

Step-by-step explanation:

To find the area of a semicircle the formula is

$\frac{\pi r^2}{2} \\=\frac{\pi 1^2}{2} \\=6.283m^2$

The square

$A=a^2=2^2=4m^2\\6.283+4=10.283m^2$

$A=wl=7*9=63m^2$ (rectangle)

$A=\frac{ab}{2} =\frac{6*4.5}{2} =13.5m^2\\63+13.5=76.5m^2$ (triangle)

5 0

4 years ago

Read 2 more answers

Consider rolling two fair dice one 3-sided the other 5-sided

Ne4ueva [31]

Since the dice are fair and the rolling are independent, each single outcome has probability 1/15. Every time we choose

$1\leq x\leq 3,\quad 1\leq y \leq 5$

We have $P(X=x)=\frac{1}{3}$ and $P(Y=y)=\frac{1}{5}$ , because the dice are fair.

Now we use the assumption of independence to claim that

$P(X=x, Y=y) = P(X=x)\cdot P(Y=y) =\dfrac{1}{3}\cdot\dfrac{1}{5} = \dfrac{1}{15}$

Now, we simply have to count in how many ways we can obtain every possible outcome for the sum. Consider the attached table: we can see that we can obtain:

2 in a unique way (1+1)
3 in two possible ways (1+2, 2+1)
4 in three possible ways
5 in three possible ways
6 in three possible ways
7 in two possible ways
8 in a unique way

This implies that the probabilities of the outcomes of $W=X+Y$ are the number of possible ways divided by 15: we can obtain 2 and 8 with probability 1/15, 3 and 7 with probability 2/15, and 4, 5 and 6 with probabilities 3/15=1/5

8 0

3 years ago