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Vilka [71]
2 years ago
10

A group of friends go ice skating. They purchase one group ticket to get in and then each person rents skates and buys a hot cho

colate.
It costs
$
10
to purchase a group ticket to get into the skating rink.
It costs
$
4
per person to rent skates.
It costs
$
1.75
per person for hot chocolate. ​​​​​​​
How many friends go ice skating if they spend a total of
$
38.75
?
Mathematics
1 answer:
Scilla [17]2 years ago
5 0

The number of friends that can go ice skating is 5.

<h3>What is the total number of friends that can go ice-skating?</h3>

The total amount spent = cost of the group ticket + total cost of the hot chocolate + total cost of renting the skates

$38.75 = $10 + $4x + $1.75x

Where x represents the total number of friends.

In order to determine the total number of friends, take the following steps:

Combine similar terms

$38.75 - $10 = $4x + $1.75x

Add similar terms

$28.75 = $5.75x

Divide both sides of the equation by 3.75

x = 5

To learn more about division, please check: brainly.com/question/13281206

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Answer:

C

Step-by-step explanation:

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7 0
3 years ago
A vertical pole 5 feet long casts a shadow of 2 feet.if at the same time a nearby tree casts a shadow of 10 feet, how tall is th
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We use the ratio of the two triangles and equate them as

h1/s1 = h2/s2

where
h1 and s1 are the height and the length of a shadow of the pole,
and the other h2 and s2 are for the tree

Identify all the given values.
5 ft / 2 ft = (h2) / 10 ft
h2 = 25 ft
Therefore the height of the tree "h2" is 25 ft.
7 0
3 years ago
What is f(−3) for the function f(a)=−2a2−5a+4?
Readme [11.4K]

Answer:

f(-3)

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f(-3) = -2(-3²) - 5(-3) + 4

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7 0
2 years ago
Read 2 more answers
12 meters in 28 seconds =? Also, How did you come to the answer?
balandron [24]
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6 0
3 years ago
A random sample of 31 charge sales showed a sample standard deviation of $50. a 90% confidence interval estimate of the populati
Marysya12 [62]
The 100(1-\alpha)\% confidence interval of a standard deviation is given by:

\sqrt{ \frac{(n-1)s^2}{\chi^2_{1- \frac{\alpha}{2} } }} \leq\sigma\leq\sqrt{ \frac{(n-1)s^2}{\chi^2_{\frac{\alpha}{2} } }}

Given a sample size of 31 charge sales, the degree of freedom is 30 and for 90% confidence interval, 

\chi^2_{1- \frac{\alpha}{2} }=43.773 \\  \\ \chi^2_{\frac{\alpha}{2} }=18.493

Therefore, the 90% confidence interval for the standard deviation is given by

\sqrt{ \frac{(31-1)50^2}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(31-1)50^2}{18.493 }} \\  \\ \Rightarrow\sqrt{ \frac{(30)2500}{43.773 }} \leq\sigma\leq\sqrt{ \frac{(30)2500}{18.493 }} \\  \\ \Rightarrow\sqrt{ \frac{75000}{43.773 }} \leq\sigma\leq\sqrt{ \frac{75000}{18.493 }} \\  \\ \Rightarrow \sqrt{1713.38} \leq\sigma\leq \sqrt{4055.59}  \\  \\ \Rightarrow41.4\leq\sigma\leq63.7
3 0
3 years ago
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