Mrs. Epps buys 500 bags of balloons for the high school prom for $250. What is the cost per bag?

Judy’s brother Sam has a collection of 96, but what are the 10 way Sam can divide his comic books equal groups

Ierofanga [76]

He can divide by 2,3,4,6,8,12,16,24,32, and 48 but

8 0

1 year ago

What is the length of segment XY?

Brilliant_brown [7]

Answer:

StartRoot 53 EndRoot units

XY = √53

Step-by-step explanation:

Choose which is point 1 and point 2 so you don't confuse the coordinates.

Point 1 (–4, 0) x₁ = –4 y₁ = 0

Point 2 (3, 2) x₂ = 3 y₂ = 2

Use the formula for the distance between two points.

$L = \sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}$

$XY = \sqrt{(3-(-4))^{2} + (2-0)^{2}}$

$XY = \sqrt{49 + 4}$

$XY = \sqrt{53}$

Therefore the line of segment XY is √53.

6 0

2 years ago

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Brainliest for the correct awnser!!! In general, when solving a radical equation with square roots, you should first isolate the

Dominik [7]

Answer:

B

Step-by-step explanation:

8 0

2 years ago

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A savings and loan association needs information concerning the checking account balances of its local customers. A random sampl

Citrus2011 [14]

Answer:

a) 98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

b) 95% confidence interval for the standard deviation.

(214.91 , 441.53)

Step-by-step explanation:

Given a size of sample 'n' =14

given mean of the sample x⁻ = $664.14

standard deviation of the sample 'S' = $297.29.

a)

98% of confidence intervals

$(x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } , x^{-}+ t_{\alpha }\frac{S}{\sqrt{n} } )$

The degrees of freedom γ=n-1 =14-1 =13

t₁₃ = 2.650 at 98% of confidence level of signification.

$(664.14- 2.650\frac{297.29}{\sqrt{14} } , 664.14+ 2.650\frac{297.29}{\sqrt{14} } )$

on calculation, we get

(664.14-210.553 , 664.14+210.553)

(453.586 , 874.693)

98% confidence interval for the true mean checking account balance for local customers.

(453.586 , 874.693)

95% of confidence intervals

$({s\sqrt{\frac{n-1}{X^{2} _{(\frac{\alpha }{2} ,n-1) } } } ,s\sqrt{\frac{n-1}{X^2_{\frac{1-\alpha }{2},n-1 } } } )$

The degrees of freedom γ=n-1 =14-1 =13

X^2_{0.05,13} =22.36 (check table)

X^2_{0.95,13} = 5.892 (check table)

$(297.29. (\sqrt{\frac{14-1}{X^2_{0.05,13} } } ),297.29(\sqrt{\frac{14-1}{X^2_{0.95,13} } } )$

$(297.29. (\sqrt{\frac{14-1}{22.36 } } ),297.29(\sqrt{\frac{14-1}{5.892 } } )$

(214.91 , 441.53)

95% confidence interval for the standard deviation.

(214.91 , 441.53)

7 0

2 years ago

Kim's business earns $10,000 per month. Kim's non-employee expenses are $3,000 per month. If Kim wants $2,000 in profit per mont

MA_775_DIABLO [31]

Answer:

Kim's business earns $10,000 per month.

Kim's non-employee expenses are $3,000 per month.

If Kim wants $2,000 in profit per month, then the maximum amount Kim can spend for employee:

10000 - 3000 - 2000 =5000

If each employee costs $1,000 per month, Kim can recruit 5000/1000 = 5 as the maximum number of employees.

Hope this helps

:)

8 0

2 years ago

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