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Artemon [7]
4 years ago
9

3. In a set of points that Milton scored in basketball games, there is an outlier.

Mathematics
1 answer:
Lemur [1.5K]4 years ago
8 0

Answer:

Less - He wouldn't be able to score as many points.

Hope this helps you out!

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4- 7f=f- 12j jrk. it ivtivivvitvittivivtvti​
lutik1710 [3]
Answer:
f=3/2j+1/2




explain
8 0
3 years ago
(Will give brainliest)
kondaur [170]

Answer: C- Fail to reject the null hypothesis. There is not enough evidence to oppose the company's claim.

<u><em>Note: I'm not sure this is correct o_O </em></u>

I used a graphing calculator and calculated the t-interval of the jellybean sample using a 99% confidence level:

<em>tInterval 9.68,1.23,125,0.99</em>

It resulted in 9.68 ± 0.287805.

Therefore, we're 99% confident that the weight number of jellybeans would lie between 9.3922 oz and 9.9678 oz.

A weight of 9.45 oz lies within this range, therefore, it is possible that the candy company's claim is true.

5 0
3 years ago
Read 2 more answers
(8 1/3) ^2 = 4 x <br><br> x =
PtichkaEL [24]

Answer:

x = 729/4 = 182.250

Step-by-step explanation:

4 0
3 years ago
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Evaluate each expression when x = 4<br> 1. 5x + 10<br> 2. 18+ x -8<br> 3. *x + 50
GaryK [48]
See photo. Hope it helps

8 0
3 years ago
Which of the following sets is a subspace of set of prime numbers P Subscript nℙn for an appropriate value of​ n? A. All polynom
goldfiish [28.3K]

Answer:

Check the explanation

Step-by-step explanation:

A. All polynomials of the form p(t) = a + bt2, where a and b are in: This means that  A is closed under scalar mult and vector addition, and includes the zero vector.

B.All polynomials of degree exactly 4, with real coefficients: what this means is that under vector addition, B isn't closed, and it does not consist of the zero vector. What it consist of  is just polynomials with degree exactly 4. Let f=x4+1f=x4+1 and let g=−x4g=−x4. Both are in B, but their sum is not, because it has degree 0.

C. All polynomials of degree at most 4, with positive coefficients: what this means is that C is not a subspace for the reason that the positive coefficients make zero vector impossible. The restriction there also makes C not closed under multiplication by the scalar −1.

So the answer is only A :D

3 0
4 years ago
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