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Oksana_A [137]
3 years ago
8

A. Does the table represent a proportional relationship?

Mathematics
1 answer:
GuDViN [60]3 years ago
5 0
B I think sorry if it’s wrong
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While on a walk in the country, you pass a field full of horses and chickens. After a quick count, you determine there are 43 he
Irina-Kira [14]

Horse = x = 1 head and 4 feet

Chicken =y = 1 head and 2 feet

Horse + chicken = x + y = 43 ( total heads)

Horse = 43 - y

4x + 2y = 122

4(43-y) + 2y = 122

Simplify:

172-4y + 2y = 122

Combine like terms

172-2y = 122

Subtract 172 from both sides

-2y = -50

Divide both sides by-2

Y = 25

X + y = 43

X + 25 = 43

X = 18

There are 18 horses and 25 chickens.

4 0
3 years ago
he owner of a local supermarket wants to estimate the difference between the average number of gallons of milk sold per day on w
stellarik [79]

Answer:

90% confidence interval is ( -149.114, -62.666   )

Step-by-step explanation:

Given the data in the question;

Sample 1                                Sample 2

x"₁ = 259.23                            x"₂ = 365.12

s₁  = 34.713                              s₂ = 48.297

n₁ = 5                                       n₂ = 10

With 90% confidence interval for μ₁ - μ₂ { using equal variance assumption }

significance level ∝ = 1 - 90% = 1 - 0.90 = 0.1

Since we are to assume that variance are equal and they are know, we will use pooled variance;

Degree of freedom DF = n₁ + n₂ - 2 = 5 + 10 - 2 = 13

Now, pooled estimate of variance will be;

S_p^2 = [ ( n₁ - 1 )s₁² + ( n₂ - 1)s₂² ] / [ ( n₁ - 1 ) + ( n₂ - 1 ) ]

we substitute

S_p^2 = [ ( 5 - 1 )(34.713)² + ( 10 - 1)(48.297)² ] / [ ( 5 - 1 ) + ( 10 - 1 ) ]

S_p^2 = [ ( 4 × 1204.9923) + ( 9 × 2332.6 ) ] / [  4 + 9 ]

S_p^2 = [ 4819.9692 + 20993.4 ] / [  13 ]

S_p^2 = 25813.3692 / 13

S_p^2 = 1985.64378

Now the Standard Error will be;

S_{x1-x2 = √[ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

S_{x1-x2 = √[ ( 1985.64378 / 5 ) + ( 1985.64378 / 10 ) ]

S_{x1-x2 = √[ 397.128756 + 198.564378 ]

S_{x1-x2 = √595.693134

S_{x1-x2 = 24.4068

Critical Value = t_{\frac{\alpha }{2}, df = t_{0.05, df=13 = 1.771  { t-table }

So,

Margin of Error E =  t_{\frac{\alpha }{2}, df × [ ( S_p^2 / n₁ ) + ( S_p^2 / n₂ ) ]

we substitute

Margin of Error E = 1.771 × 24.4068

Margin of Error E = 43.224

Point Estimate = x₁ - x₂ = 259.23 - 365.12 = -105.89

So, Limits of 90% CI will be; x₁ - x₂ ± E

Lower Limit = x₁ - x₂ - E = -105.89 - 43.224 = -149.114

Upper Limit = x₁ - x₂ - E = -105.89 + 43.224 = -62.666

Therefore, 90% confidence interval is ( -149.114, -62.666   )

3 0
3 years ago
Rearrange the formula d = m/ v for v.
balu736 [363]
V = m/v

Hope that helps!
6 0
3 years ago
17-2p=2p+5+2p solve for p
alina1380 [7]
P=2

<span>17-2p=2p+5+2p    P=2

I hope this helps.

Can you make my answer the brainliest please?
</span>
5 0
3 years ago
Read 2 more answers
How do I solve 3w+6=4w ?
Troyanec [42]
3w+6=4w \ \ \ |\hbox{subtract 4w from both sides} \\&#10;-w+6=0 \ \ \ |\hbox{subtract 6 from both sides} \\&#10;-w=-6 \ \ \ |\hbox{mutiply both sides by (-1)} \\&#10;w=6
6 0
4 years ago
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