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shtirl [24]
3 years ago
12

Juliana and Maria logged a total of 90 hours working for their dad. Juliana worked 2 hours more than three times as much as Mari

a. How many hours did Juliana work?
Mathematics
2 answers:
Mademuasel [1]3 years ago
6 0
Answer:
Juliana worked 68 hours

Explanation:
Assume the hours logged by Juliana is x
Assume the hours logged by Maria is y

We are given that:
1- Total hours logged by both is 90. This means that:
x + y = 90 
This equation can be rewritten as:
x = 90 - y ...........> equation I

2- Juliana worked 2 hours more than 3 times as much as Maria. This means that:
x = 3y + 2 ...........> equation II

Substitute with equation I in equation II and solve for y as follows:
x = 3y + 2
90 - y = 3y + 2
88 = 4y
y = 22

Substitute with y in equation I to get x as follows:
x = 90 - y
x = 90 - 22
x = 68

Based on the above:
Hours logged by Juliana = x = 68 hours
Hours logged by Maria = y = 22 hours

Hope this helps :)
Harman [31]3 years ago
5 0

Let

x---------> the hours logged by Juliana

y--------->  the hours logged by Maria

we know that

x+y=90 ---------> equation 1

x=3y+2 ---------> equation 2

Substitute equation 2 in equation 1

(3y+2)+y=90

4y+2=90

4y=90-2

4y=88

y=88/4

y=22\ hours

<u>Find the value of x</u>

x=3y+2

x=3*22+2

x=68\ hours

therefore

<u>the answer is</u>

Juliana worked 68\ hours

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An engineer is comparing voltages for two types of batteries (K and Q) using a sample of 83 type K batteries and a sample of 77
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Answer:

1. Null Hypothesis, H_0 : \mu_1 = \mu_2  {mean voltage for these two types of

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Alternate Hypothesis, H_1 : \mu_1\neq \mu_2 {mean voltage for these two types of

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2. Test Statistics value = -5.06

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Step-by-step explanation:

We are given that an engineer is comparing voltages for two types of batteries (K and Q).

where, \mu_1 = true mean voltage for type K batteries.

            \mu_2 = true mean voltage for type Q batteries.

So, <em>Null Hypothesis, </em>H_0<em> : </em>\mu_1 = \mu_2<em>  {mean voltage for these two types of </em>

<em>                                                          batteries is same}</em>

<em>Alternate Hypothesis, </em>H_1<em> : </em>\mu_1\neq \mu_2<em> {mean voltage for these two types of </em>

<em>                                                             batteries is different]</em>

The test statistics we use here will be :

                          \frac{(X_1bar-X_2bar) - (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}  } }   follows t_n__1+n_2-2

where, X_1bar = 9.29         and      X_2bar =  9.65

                s_1    = 0.374       and             s_2 =  0.518

                 n_1   = 83            and             n_2  =  77

                  s_p = \sqrt{\frac{(n_1-1)s_1^{2}+(n_2-1)s_2^{2}  }{n_1+n_2-2} } =   \sqrt{\frac{(83-1)0.374^{2}+(77-1)0.518^{2}  }{83+77-2} } = 0.45                                                   Here, we use t test statistics because we know nothing about population standard deviations.

      Test statistics = \frac{(9.29-9.65) - 0 }{0.45\sqrt{\frac{1}{83}+\frac{1}{77}  } }  follows t_1_5_8

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