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vredina [299]
3 years ago
10

if Chelsea has 11 times as many art. Brushes and they have 60 art brushes altogether how many brushes does Chelsea have

Mathematics
1 answer:
scoundrel [369]3 years ago
8 0
<span>t's call P the number of art brushes that Monique has. Chelsea has 11xP (that is 11 times as many as Monique) Together they have: P + 11xP = 12xP That means 12xP = 60 (60 is the number of art brushes they have together) In order to find how much is P, we need to divide: 60 Ă· 12 = 5 brushes P = 5 ; that means has Monique 5 art brushes. Chelsea has 11 times as many art brushes as Monique: 11 x 5 = 55 brushes</span>
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Two 5-year girls, Alyse and Jocelyn, have been training to run a 1-mile race. Alyse's 1 mile time A is approximately Normally di
tatyana61 [14]

Answer:

1.7 × 10⁻⁴

Step-by-step explanation:

The question relates to a two sample z-test for the comparison between the means of the two samples  

The null hypothesis is H₀:  μ₁ ≤ μ₂

The alternative hypothesis is Hₐ: μ₁ > μ₂

z=\dfrac{(\bar{x}_1-\bar{x}_2)-(\mu_{1}-\mu _{2} )}{\sqrt{\dfrac{\sigma_{1}^{2} }{n_{1}}-\dfrac{\sigma _{2}^{2}}{n_{2}}}}

Where;

\bar {x}_1 = 13.5

\bar {x}_2 = 12

σ₁ = 2.5

σ₂ = 1.5

We set our α level at 0.05

Therefore, our critical z = ± 1.96

For n₁ = n₂ = 23, we have;

z=\dfrac{(13.5-12)-(0)}{\sqrt{\dfrac{2.5^{2} }{23}-\dfrac{1.5^{2}}{23}}} = 3.5969

We reject the null hypothesis at α = 0.05, as our z-value, 3.5969 is larger than the critical z, 1.96 or mathematically, since 3.5969 > 1.96

Therefore, there is enough statistical evidence to suggest that Alyse time is larger than Jocelyn in a 1 mile race on a randomly select day and the probability that Alyse has a larger time than Jocelyn is 0.99983

Therefore;

The probability that Alyse has a smaller time than Jocelyn is 1 - 0.99983 = 0.00017 = 1.7 × 10⁻⁴.

8 0
3 years ago
The function f(t) = t2 + 6t − 20 represents a parabola.
Novay_Z [31]
Part A: f(t) = t² + 6t - 20
              u = t² + 6t - 20
         + 20            + 20
      u + 20 = t² + 6t
u + 20 + 9 = t² + 6t + 9
      u + 29 = t² + 3t + 3t + 9
      u + 29 = t(t) + t(3) + 3(t) + 3(3)
      u + 29 = t(t + 3) + 3(t + 3)
      u + 29 = (t + 3)(t + 3)
      u + 29 = (t + 3)²
          - 29       - 29
              u = (t + 3)² - 29

Part B: The vertex is (-3, -29). The graph shows that it is a minimum because it shows that there is a positive sign before the x²-term, making the parabola open up and has a minimum vertex of (-3, -29).
------------------------------------------------------------------------------------------------------------------
Part A: g(t) = 48.8t + 28           h(t) = -16t² + 90t + 50
            | t |   g(t)  |                          |  t  |  h(t)  |
            |-4|-167.2|                          | -4 | -566 |
            |-3|-118.4|                          | -3 | -364 |
            |-2| -69.6 |                          | -2 | -194 |
            |-1| -20.8 |                          | -1 |  -56  |
            |0 |   -28  |                          |  0  |   50  |
            |1 |  76.8 |                          |  1  |  124 |
            |2 | 125.6|                          |  2  | 166  |
            |3 | 174.4|                          |  3  | 176  |
            |4 | 223.2|                          |  4  | 154  |
The two seconds that the solution of g(t) and h(t) is located is between -1 and 4 seconds because it shows that they have two solutions, making it between -1 and 4 seconds.

Part B: The solution from Part A means that you have to find two solutions in order to know where the solutions of the two functions are located at.
4 0
3 years ago
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Lady_Fox [76]

Answer:

D.

Step-by-step explanation:

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7 0
3 years ago
Read 2 more answers
I need help with this
Flauer [41]
The given angles are complementary, therefore:

(5r + 5) + (8r -6) = 90°

13r - 1 = 90
13r - 1 + 1 = 90 + 1
13r = 91

13r/13 = 91/13

r = 7
6 0
2 years ago
Question 17 (2 points)
qwelly [4]
The answer is T
Hope this helps.
4 0
2 years ago
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