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Ostrovityanka [42]
3 years ago
5

Tickets to a local movie were sold at $3.00 fir adults and $1.5 for students. If 260 tickets were sold for a total of $675 how m

any adult tickets were sold
Mathematics
1 answer:
Alex73 [517]3 years ago
3 0

Easy all it would be is.. 260 times 675 that would be 175,500.

Yould have to just multiply. So the answer is 175,500

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If y varies directly as x and y = 24 when<br> x= 16, what is y when x = 50?
mars1129 [50]

Answer:

y = 75

Step-by-step explanation:

Given

y\ \alpha\ \ x

y=24; x =16

Required

Find y when x = 50

We have:

y\ \alpha\ \ x

Express as equation

y = kx

Solve for k

k = \frac{y}{x}

When y=24; x =16

k = \frac{24}{16}

k = \frac{3}{2}

When x = 50, we have:

y = kx

y = \frac{3}{2} * 50

y = 75

3 0
3 years ago
Exam: 05.03 Exponential and Logarithmic Functions Mid-Module Check<br><br><br> Solve 4x = 8x − 1
Fantom [35]
If 4x = 8x - 1 then 4x must be equivalent to 1. So that means x is .25 or 1/4.

Hope this helps!
(I assumed your trying to solve for x, please word the question better next time)
5 0
3 years ago
Read 2 more answers
Alex has a price of fabric that is 56 inches long and 3 inches wide. He cuts it into pieces that are each 7 inches long and 1 in
devlian [24]
Each is 7 long and the whole piece is 56 so we can make 8 down the length and each is 1 wide and the whole things is 3 so that makes 3 across. we multiply these to get 8*3= 24
3 0
3 years ago
While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem
Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = <em><u>population proportion</u></em>.

So, Null Hypothesis, H_0 : p = 0.013      {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, H_A : p > 0.013      {means that this is an unusually high number of faulty modems}

The test statistics that would be used here <u>One-sample z-test</u> for proportions;

                             T.S. =  \frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }  ~  N(0,1)

where, \hat p = sample proportion faulty modems= \frac{10}{367} = 0.027

           n = sample of modems = 367

So, <u><em>the test statistics</em></u>  =  \frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }

                                     =  2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. <u>Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.</u>

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u><em>we reject our null hypothesis</em></u>.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0
3 years ago
you work at a hair salon and get left a 35 tip on a 112 haircut and color service . What percent tip did your customer give you
levacccp [35]
31.25 percent of 112 is 35
7 0
3 years ago
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