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seropon [69]
2 years ago
5

Luis has $4.00 from his allowance. He visits an arcade where soda costs $0.75 and video games cost $0.25. Which of the following

expressions represents the amount of money Luis will have left if he buys one soda and plays "g" video games? *
5 points
4.00 - 0.25 - 0.75g
4.00 + 0.75 + 0.25g
4.00 - 0.75 - 0.25g
4.00 - 0.25 + 0.75g
Mathematics
1 answer:
nordsb [41]2 years ago
3 0

Answer: c

Step-by-step explanation:

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Two planes are the same altitude. From the airport , one plane is 50 km away in the direction of N°60 E and another is 80 km awa
djverab [1.8K]

9514 1404 393

Answer:

  78.5 km

Step-by-step explanation:

Measured at the airport, the angle between the two planes is ...

  180° -60° -50° = 70°

The law of cosines tells us the distance between the planes is ...

  d = √(50² +80² -2·50·80·cos(70°)) ≈ √6163.84 ≈ 78.5 . . . km

The planes are about 78.5 km apart.

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3 years ago
Hep asappppppppppppp
schepotkina [342]

Answer:

(6 , -1)

Step-by-step explanation:

2x+  y  = 11   ----------------(I)

y = 11 - 2x --------------------(II)

\frac{1}{2}x - 5y = 8

Multiply the whole equation by 2

2*\frac{1}{2}x -2*5y = 2*8

x - 10y = 16 --------------------(III)

Substitute y = 11- 2x in equation (III)

x - 10(11 - 2x) = 16

x - 110 + 20x = 16

21x - 110 = 16

21x = 16 +110

21x = 126

x = 126/21

x = 6

Plugin x = 6 in equation (II)

y = 11 - 2*6

y = 11 - 12

y = -1

5 0
2 years ago
If the probability of an event is 0.7 repeating​, what are the odds against the​ event
ivann1987 [24]
So the probability is the number of favorable outcomes divided by the number of total outcomes. This means that the favorable outcomes are 7/9, and the unfavorable outcomes are 2/9. The odds against the event are the unfavorable outcomes. Therefore the odds against the event is 2/9, or 0.2 repeating. Hope this helps. Feel free to ask more questions, and feel free to ask questions about my explanation.
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Answer:

Step-by-step explanation:

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3 years ago
Read 2 more answers
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
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