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Veseljchak [2.6K]
3 years ago
5

Kim and Danielle each improved their yards by planting hostas and ivy. They bought their supplies from the same store. Kim spent

$62 on 5 hostas and 4 pots of ivy. Danielle spent $144 on 8 hostas and 12 pots of ivy. What is the cost of one hosta?
Mathematics
1 answer:
svetlana [45]3 years ago
8 0

Answer:

The cost of one hosta is $6

Step-by-step explanation:

A system of linear equations is a set of two or more equations of the first degree, in which two or more unknowns are related.

Solving a system of equations consists of finding the value of each unknown so that all the equations of the system are satisfied.

In this case, the variables are:

  • x = price of a hosta .
  • y = price of a pot of ivy.

If Kim spent $62 on 5 hostas and 4 pots of ivy, t his is represented by the equation: 5*x +4*y= 62

If Danielle spent $144 on 8 hostas and 12 pots of ivy, t his is represented by the equation: 8*x +12*y=144

So, the system of equations to solve is:

\left \{ {{5*x+4*y=62} \atop {8*x+12*y=144}} \right.

One way to solve a system of equations is through the substitution method, which consists of solving or isolating one of the unknowns (for example, x) and substituting its expression in the other equation. In this way, you obtain an equation of the first degree with the other unknown, and. Once solved, you calculate the value of x by substituting the already known value of y.

Isolating y from the first equation:

y=\frac{62-5*x}{4}

and replacing this expression in the second equation you get:

8*x +12*\frac{62-5*x}{4}=144

Solving:

8*x +\frac{12}{4}*(62-5*x)=144

8*x + 3*(62-5*x)=144

8*x + 3*62 - 3*5*x=144

8*x + 186 - 15*x=144

8*x  - 15*x=144 -186

(-7)*x= -42

x= (-42)÷ (-7)

x= 6

<u><em>The cost of one hosta is $6</em></u>

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Answer:

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(b) The value of t test statistics is 1.890.

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Step-by-step explanation:

We are given that a recent study focused on the number of times men and women who live alone buy take-out dinner in a month.

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Statistic : Men      Women

The sample mean : 24.51      22.69

Sample standard deviation : 4.48    3.86

Sample size : 35    40

<em>Let </em>\mu_1<em> = mean number of times men order take-out dinners in a month.</em>

<em />\mu_2<em> = mean number of times women order take-out dinners in a month</em>

(a) So, Null Hypothesis, H_0 : \mu_1-\mu_2 = 0     {means that there is no difference in the mean number of times men and women order take-out dinners in a month}

Alternate Hypothesis, H_A : \mu_1-\mu_2\neq 0     {means that there is difference in the mean number of times men and women order take-out dinners in a month}

The test statistics that would be used here <u>Two-sample t test statistics</u> as we don't know about the population standard deviation;

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where, \bar X_1 = sample mean for men = 24.51

\bar X_2 = sample mean for women = 22.69

s_1 = sample standard deviation for men = 4.48

s_2 = sample standard deviation for women = 3.86

n_1 = sample of men = 35

n_2 = sample of women = 40

Also,  s_p=\sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} }  =  \sqrt{\frac{(35-1)\times 4.48^{2}+(40-1)\times 3.86^{2}  }{35+40-2} } = 4.16

So, <u>test statistics</u>  =  \frac{(24.51-22.69)-(0)}{4.16 \sqrt{\frac{1}{35}+\frac{1}{40}  } }  ~ t_7_3

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(b) The value of t test statistics is 1.890.

(c) Now, at 0.01 significance level the t table gives critical values of -2.651 and 2.651 at 73 degree of freedom for two-tailed test.

Since our test statistics lies within the range of critical values of t, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which <u>we fail to reject our null hypothesis</u>.

Therefore, we conclude that there is no difference in the mean number of times men and women order take-out dinners in a month.

(d) Now, the P-value of the test statistics is given by;

                     P-value = P( t_7_3 > 1.89) = 0.0331

So, P-value for two tailed test is = 2 \times 0.0331 = <u>0.0662</u>

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