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3 years ago

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Clara goes miniature golfing. She pays $7.50 for an admission ticket and $6.25 for each round she golfs. The total amount Clara

pays for admission and the number of rounds she golfs is $26.25. Which equation can be used to determine the number of rounds, x, that Clara golfs?
A - 6.25x + 7.50 = 26.25

B - G.25x - 7.50 = 26.25

C - 7.50x + 6.25 = 26.25

D - 7.50x - 6.25 = 26.25

2 answers:

Sphinxa [80]3 years ago

7 0

Answer:

6.25x + 7.50 = 26.25

Step-by-step explanation:

Clara played 3 rounds of golf.

The admission fee is = $7.50

Let the number of rounds be = x

Each round costs = $6.25

Total money Clara paid = $26.25

We can denote this in equation form as:

6.25x + 7.50 = 26.25

Now we will solve for x

6.25x = 26.25 - 7.50

=> 6.25x = 18.75

=> x = 3

Hence, Clara played 3 rounds of golf.

MrRa [10]3 years ago

4 0

Answer:

A.

Step-by-step explanation:

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Gekata [30.6K]

Answer:

9

Step-by-step explanation:

So there's 3 1's, 2 2's, 2 3's, 1 4, 2 5's, a 6, a 7, an 8, 2 0's, and no 9's.

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3 years ago

Read 2 more answers

Solve for x:<br> 4x + 10 = 34

VladimirAG [237]

Answer:

x=6

Step-by-step explanation:

4x+10=34

4x=24

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3 years ago

Read 2 more answers

1: Ken earned $176 from his part-time job this summer. He spent 25% of his money on games. He is going to donate 14 of the rema

tatiyna

The charity will receive $ 11

<h3><u>Solution:</u></h3>

Ken earned $176 from his part-time job this summer

Amount earned = $ 176

He spent 25% of his money on games

Therefore,

Money spent on games = 176 - 25 % of 176

$\text{Money spent on games } = 176 - \frac{25}{100} \times 176\\\\\text{Money spent on games } = 176 - 0.25 \times 176\\\\\text{Money spent on games } = 176 - 44\\\\\text{Money spent on games } = 132$

He is going to donate 1/4 of the remaining money to charity

Reamining amount = 176 - 132 = 44

Therefore,

$Money\ given\ to\ charity = \frac{1}{4} \times 44 = 11$

Thus he gave $ 11 to charity

8 0

3 years ago

Boxes of raisins are labeled as containing 22 ounces. Following are the weights, in the ounces, of a sample of 12 boxes. It is r

ZanzabumX [31]

Answer:

A 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

Step-by-step explanation:

We are given the weights, in the ounces, of a sample of 12 boxes below;

Weights (X): 21.88, 21.76, 22.14, 21.63, 21.81, 22.12, 21.97, 21.57, 21.75, 21.96, 22.20, 21.80.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean weight = $\frac{\sum X}{n}$ = 21.88 ounces

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = 0.201 ounces

n = sample of boxes = 12

$\mu$ = population mean weight

<em>Here for constructing a 90% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.</em>

<u>So, 90% confidence interval for the population mean, </u> $\mu$ <u> is ;</u>

P(-1.796 < $t_1_1$ < 1.796) = 0.90 {As the critical value of t at 11 degrees of

freedom are -1.796 & 1.796 with P = 5%}

P(-1.796 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 1.796) = 0.90

P( $-1.796 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

P( $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ) = 0.90

<u>90% confidence interval for</u> $\mu$ = [ $\bar X-1.796 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+1.796 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $21.88-1.796 \times {\frac{0.201}{\sqrt{12} } }$ , $21.88+1.796 \times {\frac{0.201}{\sqrt{12} } }$ ]

= [21.78, 21.98]

Therefore, a 90% confidence interval for the mean weight is [21.78 ounces, 21.98 ounces].

8 0

3 years ago

A skier is on the top of a 350 meter tall mountain looking down at a tree. The direct path from the tree to the skier is 505 met

denpristay [2]

Answer:

$43.87^{\circ}$

Step-by-step explanation:

We are given that

Height of mountain=350 m

Distance from the tree to the skier=505 m

We have to find the angle of depression from where the skier is standing to the base of the tree.

In triangle ABC

$\frac{AB}{AC}=sin\theta$

By using the formula

$\frac{Perpendicular\;side}{Hypotenuse}=sin\theta$

$\frac{350}{505}=sin\theta$

$sin^{-1}(\frac{350}{505})=\theta$

$\theta=43.87^{\circ}$

Hence, the angle of depression from where the skier is standing to the base of the tree= $43.87^{\circ}$

6 0

3 years ago