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Ray Of Light [21]
2 years ago
11

Suppose there are 100 cookies in a jar. if 30 are chocolate chip, 40 are oatmeal raisin, 20 are peanut butter, and 10 are sugar

cookies, what are the odds of randomly selecting either an oatmeal raisin or sugar cookie from the jar?
Mathematics
1 answer:
lubasha [3.4K]2 years ago
3 0
If there's 100 cookies, of which there are 40 oatmeal raisin and 10 sugar cookies, there are 50 out of 100 cookies which you want. Hence by chance, there is a 50/100, or 1 in 2 chance of you obtaining an oatmeal raisin or a sugar cookie.
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Solve for the equation -2y+y-3 = 6
pashok25 [27]

Answer: y = -9

Step-by-step explanation:

(-2y) + y - 3 = 6

(-y) - 3 = 6

-y = 9

y = -9

7 0
2 years ago
Sue played four games of golf for these games her modal score was 98 and her mean score was 100 her range of score was 10 what w
earnstyle [38]

Answer : Remaining two observation becomes 97 and 107.

Explanation :

Since we have given that

Mean = 100

Modal value = 98

Range = 10

As we know that ,

Range = Highest-Lowest

Let highest observation be x

Let lowest observation be y

So equation becomes x-y=10 ----equation 1

So, observation becomes

x,98,98,y

Now, we use the formula of mean i.e.

Mean = \frac{\text{Sum of observation}}{\text{N.of observaton}}

So, mean =\frac{x=98=98+y}{4}=400\\\frac{196+x+y}{4}=100\\x+y=400-196\\x-y=204

So our 2nd equation becomes

x+y=204

On using elimination method of system of linear equation on these two equation we get,

x=97

and

x+y=204\\y=204-x\\y=204-97\\y=107

Hence , remaining two observation becomes 97 and 107.

8 0
1 year ago
I see that the amount i owe has been reduced from $90.00 to $75.00."
NikAS [45]
Then your amount was reduced to $15 less
4 0
2 years ago
Read 2 more answers
The weights of certain machine components are normally distributed with a mean of 8.01 g and a standard deviation of 0.06 g. Fin
Free_Kalibri [48]

Answer:

Option D) 7.90 g and 8.12 g

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 8.01 g

Standard Deviation, σ = 0.06 g

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

Formula:

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

We have to find the value of x such that the probability is 0.03

P(X > x)  

P( X > x) = P( z > \displaystyle\frac{x - 8.01}{0.06})=0.03  

= 1 -P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.03  

=P( z \leq \displaystyle\frac{x - 8.01}{0.06})=0.97  

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

\displaystyle\frac{x - 8.01}{0.06} = 1.881\\\\x = 8.12  

Thus, 8.17 g separates the top 3% of the weights.

P(X < x)  

P( X < x) = P( z < \displaystyle\frac{x - 8.01}{0.06})=0.03  

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

\displaystyle\frac{x - 8.01}{0.06} = -1.881\\\\x = 7.90  

Thus, 7.90 separates the bottom 3% of the weights.

Thus, the correct answer is

Option D) 7.90 g and 8.12 g

7 0
2 years ago
Solve X-9+2wx=y for x
maxonik [38]
X - 9 + 2wx = y Add 9 to both sides x + 2wx = y + 9 Factor out the x x (1 + 2w) = y + 9 Divide both sides by (1 + 2w) x = (y + 9) / (1 + 2w)
8 0
1 year ago
Read 2 more answers
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