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kherson [118]
2 years ago
5

I'd x+2a=4x , then a is equivalent to

Mathematics
1 answer:
STALIN [3.7K]2 years ago
4 0
<span> x+2a=4x
2a = 3x
a = 3x/2
a =1.5x</span>
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A standard deck has 52 cards consisting of 26 black and 26 red cards. Three cards are dealt from a shuffled deck without replace
Allisa [31]

They aren't independent since the probability uses all the cards in the deck

So at the first deal we have the chance of 26/52 of getting a red card, at the second deal we have the chance of a 25/51 of getting another red card, so they aren't independent

8 0
2 years ago
Mrs smith bought a.box of 10 cupcake she ate 30% of the cupcake how many did she eat
iogann1982 [59]

Answer:

3 Cupcakes

Step-by-step explanation:

First, multiply the percentage, 30, times the number of cupcakes, 10, ad you get 300. Then, divide that by 100, and oyu get your answer, 3 cupcakes.

*Or since this form of equation is in a simpler form of 10 cupcakes, just get rid of the percentage sign and the 0 in 30%. I hope this helps :)

4 0
2 years ago
Find each of the following products of conjugate pairs.See if you can work out a pattern
yKpoI14uk [10]

Answer: is 0

Step-by-step explanation:

4 0
3 years ago
Please help me on this
IrinaVladis [17]

Answer:

Step-by-step explanation:

coordinate of O(2,1),Y(3,-2),T(-3,-3)

let the reflected points are O',Y',T'

distance of O from y=2 is 1 unit.

∴ ordinate of O' =1+2(1)=3

or O' is (2,3)

distance of Y from y=2 is2+2=4

ordinate of Y'=-2+2(4)=6

so coordinate of Y' is (3,6)

distance of T from y=2 is (2+3)=5

so ordinate of T' =-3+2(5)=7

coordinate of T' is(-3,7)

3 0
2 years ago
After a college football team once again lost a game to their archrival, the alumni association conducted a survey to see if alu
valentina_108 [34]

Answer:

P-value for this hypothesis test is 0.00175.

Step-by-step explanation:

We are given that the alumni association conducted a survey to see if alumni were in favor of firing the coach.

A simple random sample of 100 alumni from the population of all living alumni was taken. Sixty-four of the alumni in the sample were in favor of firing the coach.

<u><em>Let p = proportion of all living alumni who favored firing the coach</em></u>

SO, Null Hypothesis, H_0 : p = 0.50   {means that the majority of alumni are not in favor of firing the coach}

Alternate Hypothesis, H_A : p > 0.50   {means that the majority of alumni are in favor of firing the coach}

The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;

                                  T.S.  = \frac{\hat p-p}{{\sqrt{\frac{\hat p(1-\hat p)}{n} } } } }  ~ N(0,1)

where, \hat p  = sample proportion of the alumni in the sample who were in favor of firing the coach = \frac{64}{100} = 0.64

            n = sample of alumni = 100

So, <em><u>test statistics</u></em>  =  \frac{0.64-0.50}{{\sqrt{\frac{0.64(1-0.64)}{100} } } } }

                               =  2.92

<u>Now, P-value of the hypothesis test is given by ;</u>

         P-value = P(Z > 2.92) = 1 - P(Z \leq 2.92)

                                             = 1 - 0.99825 = 0.00175

Therefore, the P-value for this hypothesis test is 0.00175.

4 0
3 years ago
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