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SSSSS [86.1K]
3 years ago
5

Ken watches a marching band. He sees 2 rows of flute players. Six people are in each row. He sees 8 trombone players. How many f

lute or trombone players does Ken see
Mathematics
1 answer:
sukhopar [10]3 years ago
6 0
Twelve flute players.

Eight trombone players.

20 total of flute and trombone players.
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Is (1, 2) a solution to this system of equations? 10x + 4y = 18 x + 9y = 19
Arlecino [84]

Answer:

Yes

Step-by-step explanation:

The intersection point of 10x+4y=18 and x+9y=19 is (1,2) on a graph.

3 0
3 years ago
Select the correct answer.
Oduvanchick [21]

Answer:

I got B too so I sure could be right answer but if it's not then I so sorry. B. 6

8 0
2 years ago
Read 2 more answers
Helppppppppppp with 3 & 4!!!!!!!!!!!!!!!!!!!!!!!
Natalka [10]

Answer:

3 is b

4 is also b

Step-by-step explanation:

6 0
3 years ago
Can I get help on this please? I think the answer might be (A) but i'm not sure.
Alex Ar [27]
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4 0
3 years ago
It is desired to compare the hourly rate of an entry-level job in two fast-food chains. Eight locations for each chain are rando
notka56 [123]

Answer:

It can be concluded that at 5% significance level that there is no difference in the amount paid by chain A and chain B for the job under consideration

Step by Step Solution:

The given data are;

Chain A 4.25, 4.75, 3.80, 4.50, 3.90, 5.00, 4.00, 3.80

Chain B 4.60, 4.65, 3.85, 4.00, 4.80, 4.00, 4.50, 3.65

Using the functions of Microsoft Excel, we get;

The mean hourly rate for fast-food Chain A, \overline x_1 = 4.25

The standard deviation hourly rate for fast-food Chain A, s₁ = 0.457478

The mean hourly rate for fast-food Chain B, \overline x_2 = 4.25625

The standard deviation hourly rate for fast-food Chain B, s₂ = 0.429649

The significance level, α = 5%

The null hypothesis, H₀:  \overline x_1 = \overline x_2

The alternative hypothesis, Hₐ:  \overline x_1 ≠ \overline x_2

The pooled variance, S_p^2, is given as follows;

S_p^2 = \dfrac{s_1^2 \cdot (n_1 - 1) + s_2^2\cdot (n_2-1)}{(n_1 - 1)+ (n_2 -1)}

Therefore, we have;

S_p^2 = \dfrac{0.457478^2 \cdot (8 - 1) + 0.429649^2\cdot (8-1)}{(8 - 1)+ (8 -1)} \approx 0.19682

The test statistic is given as follows;

t=\dfrac{(\bar{x}_{1}-\bar{x}_{2})}{\sqrt{S_{p}^{2} \cdot \left(\dfrac{1 }{n_{1}}+\dfrac{1}{n_{2}}\right)}}

Therefore, we have;

t=\dfrac{(4.25-4.25625)}{\sqrt{0.19682 \times \left(\dfrac{1 }{8}+\dfrac{1}{8}\right)}} \approx -0.028176

The degrees of freedom, df = n₁ + n₂ - 2 = 8 + 8 - 2 = 14

At 5% significance level, the critical t = 2.145

Therefore, given that the absolute value of the test statistic is less than the critical 't', we fail to reject the null hypothesis and it can be concluded that at 5% significance level that chain A pays the same as chain B for the job under consideration

3 0
2 years ago
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