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sasho [114]
3 years ago
14

Jenna has a bag that contains 7 red marbles and 25 blue marbles. She selects a marble at random, and then, without replacing the

first one, selects another marble at random. What is the probability that Jenna selects a blue marble and then a red marble? Round your answer to the nearest percent. P(blue and red) ≈ %
Mathematics
1 answer:
slava [35]3 years ago
5 0
17.6%=D so go get 100% bud
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kiruha [24]

Answer:

5 thirds

Step-by-step explanation:

Make 1 2/3 into a improper fraction. Now you have 5/3. 5/3-(1/3*5)=0

Therefore, 5x1/3=5/3=1 2/3

7 0
3 years ago
Read 2 more answers
Alex purchased 3 packs of baseball cards that cost $2.50 each. He also purchased 2 packs of card protectors that cost $3.00 each
vampirchik [111]

Answer:

6.50

Step-by-step explanation:

You have to add/multiply 2.50 by 3 to get the total cost of the cards. 2.50*3=7.5. then, you need to add/multiply 3.00 by 2 to get 6. 7.5+6= 13.5. 20-13.5=6.50.

the change was $6.50

6 0
3 years ago
The number y of hits a new website receives each month can be modeled by y = 4070ekt, where t represents the number of months th
makvit [3.9K]

Answer:

<u><em>k = 0.2645</em></u>

Step-by-step explanation:

Given model:

y = 4070 e^{kt}

y = no. of hits website received = 9000 (in 3rd month)

t= no. of months website has been operational = 3

put in the above equation:

9000 = 4070  e^{3k}

\frac{9000}{4070} = e^{3k}

\frac{900}{407}=e^{3k}

<u><em>Taking natural logarithm on both sides, we get:</em></u>

ln\frac{900}{407}=ln(e^{3k})

ln\frac{900}{407}= 3k ln<em>e</em>

ln<em>e</em>=1

ln \frac{900}{407}= 3k

or k = \frac{1}{3}ln\frac{900}{(407)}

k =\frac{1}{3}(0.7936)

<em>k = 0.2645</em>

<em />

7 0
3 years ago
How to solve the 4a.plz
vladimir1956 [14]
(4xy-2y^2)+2y
4-2=2....x will remain y^1-y^2=y^-1
2xy^-1+2y
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3 0
2 years ago
You are interested in the relationship between leisure activities and mood. One group of participants watches at least 2 hours o
Romashka [77]

Answer:

The most appropriate statistical test to use to compare the mood scores from the different groups is independent sample t-test.

Step-by-step explanation:

The Independent Samples t-test examines the means of two independent groups to see if statistical evidence exists to show that the related population means differ significantly.

The Independent Samples t-test is also known as Independent t-test, Independent Two-sample t-test, and among others.

It should be note that only two (and only two) groups can be compared using the Independent Samples t-test. It is not possible to use it to make comparisons between more than two groups.

Therefore, the most appropriate statistical test to use to compare the mood scores from the different groups is independent sample t-test.

3 0
3 years ago
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