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SOVA2 [1]
3 years ago
10

Average movie prices in the United States​ are, in​ general, lower than in other countries. It would cost ​$81.73 to buy three t

ickets in Japan plus two tickets in Switzerland. Three tickets in Switzerland plus two tickets in Japan would cost ​$77.62. How much does an average movie ticket cost in each of these​ countries? In Japan the average cost is ​$ nothing. In Switzerland the average cost is ​$ nothing.
Mathematics
2 answers:
Vikentia [17]3 years ago
6 0
Add 81.73 +77.62 and it will equal 159.35 so your answer is 159.35
stich3 [128]3 years ago
4 0

Answer:

Step-by-step explanation:

In japan the average cost is  up from $.1200 in some change to

in Switzerland the average cost is 15.00 and some change

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Salma made 6 identical necklaces, each having beads and a pendant. The total cost of the beads and pendants for all 6 necklaces
IRINA_888 [86]

Answer:

$3.30

Step-by-step explanation:

2.60*6= $15.60 beads per necklace

35.40-15.60= $19.80 total pendant

19.80/6= $3.30 per pendant

8 0
2 years ago
Read 2 more answers
Find the missing value <br> 3/8 +1/?=5/8<br> 3/4-1/?=1/4<br> ?/6 + 1/8= 23/24
steposvetlana [31]
1st one : is 1/4 ( subtract 3/8 from both sides and simplify 5/8 - 3/8 to 1/4. )

2nd one : is 1/2 ( subtract 3/4 from both sides, simplify 1/4 - 3/4 to 1/-2, and then multiply both sides by -1 )

3rd one : is 5/6 ( subtract 1/8 from both sides and simplify 23/24 - 1/8 to 5/6).
6 0
3 years ago
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1/6d + 2/3 = 1/4(d-2)
xxMikexx [17]
D = 14

1/6d + 2/3 = 1/4d - 1/2
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4 0
3 years ago
Which expression should you simplify to find the 90% confidence interval,
Alex787 [66]

Answer:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287

0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

Step-by-step explanation:

The estimated proportion of interest is \hat p=0.45

We need to find a critical value for the confidence interval using the normla standard distributon. For this case we have 95% of confidence, then the significance level would be given by \alpha=1-0.95=0.05 and \alpha/2 =0.025. And the critical value is:

z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96

The confidence interval for the true population proportion is interest is given by this formula:  

\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}

Replacing the values provided we got:

0.45 - 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.287

0.45 + 1.96\sqrt{\frac{0.45(1-0.45)}{36}}=0.613

And we can conclude at 95% of confidence that the true proportion of interest for this case is between 0.287 and 0.613

8 0
3 years ago
Which of the following graphs represents a function?
gayaneshka [121]

Answer:

the second one

hope this helps

8 0
3 years ago
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