TODAY, Kyle has 11 pounds to lose to get to his goal weight. Tuesday he was 196 lbs. TODAY, he is down 3 lbs.

The table below shows the number of marbles of different colors in a bag:

noname [10]

There are a total of 5+8+2=15 marbles. There's a 5/15 chance of drawing a red marble in the first trial. Then, there are 4 red marbles and 14 total marbles left, so afterwards, there's a probability of 4/14 that the second marble is red. The probability overall is found by multiplying these together, so the first answer is the best.

There are 2+6+7=15 cakes. There's a 6/15 chance that the first cake selected will be a pineapple cake, and then a 2/14 chance that the second cake selected will be a strawberry cake, by the same logic as above. This is equal to 12/210. The first answer is the best option.

3 0

3 years ago

Read 2 more answers

NEED ANSWER ASAP!!! Do the data in this table show a function? If you switch the input and the output values is it a function?

ser-zykov [4K]

Answer:

no

Step-by-step explanation:

no input can have diffrent outputs if you change the input values then yes it is a function hope this helps god bless

4 0

2 years ago

When you add 18 to 1/4 of a number, you get the number itself.

Lerok [7]

So if 18 plus 1/4 of the number is the number the means 18 is 3/4 of the number so 18 divided by 3 is 6 so 1/4 = 6 now add 6 to 18 to get 4/4 of the number which is 24 number = 24

8 0

3 years ago

In which tables does y vary directly with x? Check all that apply

ratelena [41]

Answer:

first three

Step-by-step explanation:

go directly to the x- value ed2020

5 0

2 years ago

Help please solve<br> <img src="https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7B6x%5E5%2B11x%5E4-11x-6%7D%7B%282x%5E2-3x%2B1

Shkiper50 [21]

Answer:

$\displaystyle -\frac{1}{2} \leq x < 1$

Step-by-step explanation:

<u>Inequalities</u>

They relate one or more variables with comparison operators other than the equality.

We must find the set of values for x that make the expression stand

$\displaystyle \frac{6x^5+11x^4-11x-6}{(2x^2-3x+1)^2} \leq 0$

The roots of numerator can be found by trial and error. The only real roots are x=1 and x=-1/2.

The roots of the denominator are easy to find since it's a second-degree polynomial: x=1, x=1/2. Hence, the given expression can be factored as

$\displaystyle \frac{(x-1)(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)^2(x-\frac{1}{2})^2} \leq 0$

Simplifying by x-1 and taking x=1 out of the possible solutions:

$\displaystyle \frac{(x+\frac{1}{2})(6x^3+14x^2+10x+12)}{(x-1)(x-\frac{1}{2})^2} \leq 0$

We need to find the values of x that make the expression less or equal to 0, i.e. negative or zero. The expressions

$(6x^3+14x^2+10x+12)$

is always positive and doesn't affect the result. It can be neglected. The expression

$(x-\frac{1}{2})^2$

can be 0 or positive. We exclude the value x=1/2 from the solution and neglect the expression as being always positive. This leads to analyze the remaining expression

$\displaystyle \frac{(x+\frac{1}{2})}{(x-1)} \leq 0$