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solniwko [45]
2 years ago
8

When Katie was visiting her grandpas farm she saw the farm only raised hands and pigs. Katie counted 32 heads and 100 feet in th

e backyard. How many hands and pigs were there in the backyard
Mathematics
1 answer:
QveST [7]2 years ago
8 0

There were 14 hens and 18 pigs in the backyard

Step-by-step explanation:

Katie was visiting her grandpas farm

  • She saw only hens and pigs
  • Katie counted 32 heads and 100 feet in the backyard

We need to find how many hens and pigs in the backyard

Assume that there are x hens and y pigs in the backyard

∵ There are x hens and y pigs in the backyard

∵ There are 32 heads

∴ x + y = 32 ⇒ (1)

∵ Each hen has 2 feet

∵ Each pig has 4 feet

∵ There are 100 feet

∴ 2x + 4y = 100 ⇒ (2)

Let us solve the system of equations to find how many hens and pigs

Multiply equation (1) by -2 to eliminate x

∴ -2x - 2y = -64 ⇒ (3)

- Add equations (2) and (3)

∴ 2y = 36

- Divide both side by 2

∴ y = 18

Substitute the value of y in equation (1) to find the value of x

∵ x + 18 = 32

- Subtract 18 from both sides

∴ x = 14

There were 14 hens and 18 pigs in the backyard

Learn more:

You can learn more about the system of equations in brainly.com/question/2115716

#learnwithBrainly

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It takes Daphne 25 minutes to assemble a model plane. During her work day Daphne takes 30 minutes for lunch and one 15 minute br
natka813 [3]

Answer:

<h2>y=25p-45</h2>

Step-by-step explanation:

We first of all start by cumulating all the time she spent for break

lunch= 30min

break= 15min

total break= 45min

also the time taken to assemble one model plane 25min

let the number of hours be p

and the total number of model plane be y

The model is

y=25p-45

7 0
2 years ago
Five days a week, you carpool with 3 co-workers and take turns driving each week. It is 14 miles from your home to your office.
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you will save 10 gallons

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If a certain number is increased by 5 one half of the result is three-fifths of the excess of 61 over the number find the number
PSYCHO15rus [73]

Answer:

= 391/11

Let say number = N

a certain number is increased by 5

= N + 5

,one-half of the result

= (N + 5)/2

Three -fifths of the excess of 61 over the number.

= (3/5)(61 - N)

Equating both

(N + 5)/2  = (3/5)(61 - N)

multiplying by 10 both sides

=> 5(N + 5) = 6(61 - N)

=> 5N + 25 = 366 - 6N

=> 11N = 391

=> N = 391/11

Step-by-step explanation:

8 0
1 year ago
The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer se
tankabanditka [31]

Answer:

A)H0: μ1 − μ2 = 0

Ha: μ1 − μ2 ≠ 0

(b)  Means

Company A___50.6___ min.

Company B___52.75___ min.

c)The t-value is -0.30107.

The p-value is 0 .764815.

C) Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Step-by-step explanation:

A)H0: μ1 − μ2 = 0

i.e there is no difference between the means of delayed flight for two different airlines

Ha: μ1 − μ2 ≠ 0

i.e there is a difference between the means of delayed flight for two different airlines

(b)

Company A___50.6___ min.

Company B___52.75___ min.

Mean of Company A = x`1= ∑x/n =34+ 59+ 43+ 30+ 3+ 32+ 42+ 85+ 30+ 48+ 110+ 50+ 10+ 26+ 70+ 52+ 83+ 78+ 27+ 70+ 27+ 90+ 38+ 52+ 76/25

= 1265/25= 50.6

Mean of Company B = x`2= ∑x/n =

=46+ 63+ 43+ 33+ 65+ 104+ 45+ 27+ 39+ 84+ 75+ 44+ 34+ 51+ 63+ 42+ 34+ 34+ 65+ 64/20

= 1055/20= 52.75

<u><em>Difference Scores Calculations</em></u>

Company A

Sample size for Company A= n1= 25

Degrees of freedom for company A= df1 = n1 - 1 = 25 - 1 = 24

Mean for Company A= x`1=  50.6

Total Squared Difference (x-x`1) for Company A= SS1: 16938

s21 = SS1/(n1 - 1) = 16938/(25-1) = 705.75

Company B

Sample size for Company B= n1= 20

Degrees of freedom for company B= df2 = n2 - 1 = 20 - 1 = 19

Mean for Company B= x`2=  52.75

Total Squared Difference (x-x`2) for Company B= SS2=7427.75

s22 = SS2/(n2 - 1) = 7427.75/(20-1) = 390.93

<u><em>T-value Calculation</em></u>

<u><em>Pooled Variance= Sp²</em></u>

Sp² = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)

Sp²= ((24/43) * 705.75) + ((19/43) * 390.93) = 566.65

s2x`1 = s2p/n1 = 566.65/25 = 22.67

s2x`2 = s2p/n2 = 566.65/20 = 28.33

t = (x`1 - x`2)/√(s2x`1 + s2x`2) = -2.15/√51 = -0.3

The t-value is -0.30107.

The total degrees of freedom is = n1+n2- 2= 25+20-2=43

The critical region for two tailed test at significance level ∝ =0.05 is

t(0.025) (43) = t > ±2.017

Since the calculated value of t=  -0.30107.  does not fall in the critical region t > ±2.017, null hypothesis is not rejected that is there is no difference between the means of delayed flight for two different airlines.

The p-value is 0 .764815. The result is not significant at p < 0.05.

7 0
1 year ago
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