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olya-2409 [2.1K]

3 years ago

9

Xander rolls two 6-sided number cubes. What is the theoretical probability of rolling an even number on the first cube and a num

ber divisible by 3 on the
second cube?

1 answer:

NeX [460]3 years ago

7 0

Answer:

1/6

Step-by-step explanation:

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HELP NEEDED PLEASE HELP

vova2212 [387]

Answer:

<em><u>3</u></em><em><u>0</u></em><em><u>0</u></em>

it is given that 12% were rotten

12% = 36

then let total no of apple be X

then , x*12/100 = 36

x= 36*100/12

x = 300

hence total apple were. = 300

hope it helps

6 0

3 years ago

Read 2 more answers

A student expressed the sum of 2 whole numbers as 5 times (8+3).What are the 2 whole numbers?

Marta_Voda [28]

Well if the two numbers are equal to 5x(8+3).

Lets say that these two numbers are X and Y,

So that would be X + Y = 5(8+3)

which means X + Y = 55

You could write that as a fuctions Y = 55 - X

Now by using the graph you get and the X values of [0,55] ( This means every value from 0 to 55 ), you will get a X and Y value that if you added up together will give you the value of 8(5+33)

An examples for the answers would be

(0,55) (1,54) (2,53) (3,52) (4,51) (5,50) (6,49) ... it goes on until x reaches 55.

8 0

3 years ago

A random sample of 12 recent college graduates reported an average starting salary of $54,000 with a standard deviation of $6,00

Marianna [84]

Answer: a.) $50188 to $57812

Step-by-step explanation: <u>Confidence</u> <u>Interval</u> (CI) is an interval of values in which we are confident the true mean is in.

The interval is calculated as

x ± $z\frac{s}{\sqrt{n} }$

a. For a 95% CI, z-value is 1.96.

Solving:

54,000 ± $1.96.\frac{6000}{\sqrt{12} }$

54,000 ± $1.96\frac{6000}{3.464}$

54,000 ± 1.96*1732.102

54,000 ± 3395

This means the interval is

50605 < μ < 57395

<u>With a 95% confidence interval, the mean starting salary of college graduates is between 50605 and 57395 or </u><u>from 50188 to 57812$.</u>

<u />

b. The mean starting salary for college students in 2017 is $50,516, which is in the confidence interval. Therefore, since we 95% sure the real mean is between 50188 and 57812, there was no significant change since 2017.

4 0

3 years ago

Karen finished watching a movie at 1:00 PM. The movie lasted 1 hour 38 minutes. at what time Karen started watching the movie. -

Andrew [12]

The correct answer would be 11:32am. i hope this helps!

6 0

3 years ago

Read 2 more answers

For every integer k from 1 to 10, inclusive the "k"th term of a certain sequence is given by (−1)(k+1)∗(12k). If T is the sum of

Katena32 [7]

Answer:

Option D. is the correct option.

Step-by-step explanation:

In this question expression that represents the kth term of a certain sequence is not written properly.

The expression is $(-1)^{k+1}(\frac{1}{2^{k}})$ .

We have to find the sum of first 10 terms of the infinite sequence represented by the expression given as $(-1)^{k+1}(\frac{1}{2^{k}})$ .

where k is from 1 to 10.

By the given expression sequence will be $\frac{1}{2},\frac{(-1)}{4},\frac{1}{8}.......$

In this sequence first term "a" = $\frac{1}{2}$

and common ratio in each successive term to the previous term is 'r' = $\frac{\frac{(-1)}{4}}{\frac{1}{2} }$

r = $-\frac{1}{2}$

Since the sequence is infinite and the formula to calculate the sum is represented by

$S=\frac{a}{1-r}$ [Here r is less than 1]

$S=\frac{\frac{1}{2} }{1+\frac{1}{2}}$

$S=\frac{\frac{1}{2}}{\frac{3}{2} }$

S = $\frac{1}{3}$

Now we are sure that the sum of infinite terms is $\frac{1}{3}$ .

Therefore, sum of 10 terms will not exceed $\frac{1}{3}$

Now sum of first two terms = $\frac{1}{2}-\frac{1}{4}=\frac{1}{4}$

Now we are sure that sum of first 10 terms lie between $\frac{1}{4}$ and $\frac{1}{3}$

Since $\frac{1}{2}>\frac{1}{3}$

Therefore, Sum of first 10 terms will lie between $\frac{1}{4}$ and $\frac{1}{2}$ .

Option D will be the answer.

3 0

3 years ago