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Lena [83]
2 years ago
12

If ABCD Is a parallelogram, then what is the m

Mathematics
1 answer:
Degger [83]2 years ago
6 0

Step-by-step explanation:

the given shape is a parallelogram so, adjacent angles are supplementary.

so,

=》3y - 15° + 2y = 180°

=》5y = 180° + 15°

=》y = 195° ÷ 5

=》y = 39°

now , Angle A = Angle C

( they are opposite angles of a parallelogram )

=》Angle A

=》 2 × y

=》2 × 39°

=》78°

hence, Angle A = 78°

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What is the value of log 43? Use the calculator. Round your answer to the nearest tenth
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Answer:

1.6

Step-by-step explanation:

The value of log 43 to the nearest tenth is <em>1</em><em>.</em><em>6</em>

3 0
3 years ago
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Which ordered pair represents the y-intercept of the graph of the equation y = -3x - 9?
Mnenie [13.5K]
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Equation
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y = -3x - 9

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Option 1
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If I substitute x = -9, I should get y = 0

When x = -9
y = -3 (-9) - 9
= 18 (I did not get 0, wrong)


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Option 2
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If I substitute x = -3, I should get y = 0
y = -3(-3) - 9
y = 9 - 9 
y = 0 (Yes, I got 0, correct)


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Option 3
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If I substitute x = 0, I should get y = -3
y = -3 (0) - 9
y = 0 - 9
y = -9 (I did not get -3, wrong)


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Option 4
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If I substitute x = 0, I should get y = -9
y = -3 (0) - 9
y = 0 - 9
y = -9 (Yes, I got -9, correct)


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Answer: (-3, 0) and (0, 9) are ordered pairs of the equation (Answer B, D)
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3 years ago
J.J.Bean sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the internet.
melamori03 [73]

Answer:

99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

Step-by-step explanation:

We are given that a random sample of 16 sales receipts for mail-order sales results in a mean sale amount of $74.50 with a standard deviation of $17.25.

A random sample of 9 sales receipts for internet sales results in a mean sale amount of $84.40 with a standard deviation of $21.25.

The pivotal quantity that will be used for constructing 99% confidence interval for true mean difference is given by;

                      P.Q.  =  \frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \times \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }  ~ t__n_1_+_n_2_-_2

where, \bar X_1 = sample mean for mail-order sales = $74.50

\bar X_2 = sample mean for internet sales = $84.40

s_1 = sample standard deviation for mail-order purchases = $17.25

s_2 = sample standard deviation for internet purchases = $21.25

n_1 = sample of sales receipts for mail-order purchases = 16

n_2 = sample of sales receipts for internet purchases = 9

Also,  s_p =\sqrt{\frac{(n_1-1)\times s_1^{2}+(n_2-1)\times s_2^{2} }{n_1+n_2-2} }  =  \sqrt{\frac{(16-1)\times 17.25^{2}+(9-1)\times 21.25^{2} }{16+9-2} } = 18.74

The true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is represented by (\mu_1-\mu_2).

Now, 99% confidence interval for (\mu_1-\mu_2) is given by;

             = (\bar X_1-\bar X_2) \pm t_(_\frac{\alpha}{2}_)  \times s_p \times \sqrt{\frac{1}{n_1} +\frac{1}{n_2}}

Here, the critical value of t at 0.5% level of significance and 23 degrees of freedom is given as 2.807.

          = (74.50-84.40) \pm (2.807  \times 18.74 \times \sqrt{\frac{1}{16} +\frac{1}{9}})

          = [$-31.82 , $12.02]

Hence, 99% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases is [$(-31.82) , $12.02].

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