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skad [1K]
3 years ago
6

What is the system of linear equations graphed below?

Mathematics
2 answers:
pochemuha3 years ago
6 0

Answer:

It's:

y = -\frac{1}{2}x + 1

y = -2x - 4

Hope this helps!

inn [45]3 years ago
4 0
Answer
y=-2x+1
y=-1/2x-4
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Sherry has taken 10 quizzes. Her combined total for the 10 quizzes is 83 points. Nine of her quiz scores are shown on the plot.
castortr0y [4]

Answer:

9

Step-by-step explanation:

4 0
3 years ago
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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].

5 0
3 years ago
Solve the inequality.<br> 8-2x &lt; -x + 9
mezya [45]

Answer:

x > -1

Step-by-step explanation:

8-2x < -x+9\\8-2x-8 < -x+9-8\\-2x < -x+1\\-2x+x < -x+1+x\\-x < 1\\\left(-x\right)\left(-1\right) > 1\cdot \left(-1\right)\\x > -1

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2 years ago
Do question a,b,c,d plz
julsineya [31]
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B) 164.64
C) 57.1
D) 145173.954289 or <span>381.017</span><span>​<span>2

A good calculator website is </span></span>https://www.cymath.com/
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What is he least common multiple of 5 and 6
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Answer:

30.

Step-by-step explanation:

6 = 2 * 3

5 = 5

LCM = 2 * 3 * 5 = 30.

7 0
3 years ago
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