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Wewaii [24]
3 years ago
9

The goal of a benefit concert is to collect 50,000 cans of food. Food is collected at each entrance to the concert. Entrance A,

15,682 cans, Entrance B, 18,940 cans, Entrance C, 16,114 cans, Entrance D, 12,468 cans. Did they collect more or fewer cans than desired? How many more or fewer?
Mathematics
2 answers:
Karo-lina-s [1.5K]3 years ago
5 0

Answer: Here you need to add the total cans that they collected in each entrance; it is 15682 cans + 18940 cans + 16115 cans + 12468 cans = 63205 total cans collected.

So at the end they collected more cans than desired, they wanted 50000 and got 63205, so if you do a difference, 63205 - 50000 = 13205

We can see that they collected 13205 more cans than desired.

Arada [10]3 years ago
4 0
In order to do this you add 15,682+18,940+16,114+12,468= 63,234
And that is larger than 50,000 so it is more.
63,324-50,000= 13,234
~JZ
Hope it helps

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olya-2409 [2.1K]
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Read 2 more answers
A high school principal wishes to estimate how well his students are doing in math. Using 40 randomly chosen tests, he finds tha
ollegr [7]

Answer:

99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Step-by-step explanation:

We are given that a high school principal wishes to estimate how well his students are doing in math.

Using 40 randomly chosen tests, he finds that 77% of them received a passing grade.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion is given by;

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

where, \hat p = sample proportion of students received a passing grade = 77%

           n = sample of tests = 40

           p = population proportion

<em>Here for constructing 99% confidence interval we have used One-sample z proportion test statistics.</em>

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                           level of significance are -2.5758 & 2.5758}  

P(-2.5758 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.5758) = 0.99

P( -2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

 = [ 0.77-2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } , 0.77+2.5758 \times {\sqrt{\frac{0.77(1-0.77)}{40} } } ]

 = [0.5986 , 0.9414]

Therefore, 99% confidence interval for the population proportion of passing test scores is [0.5986 , 0.9414].

Lower bound of interval = 0.5986

Upper bound of interval = 0.9414

6 0
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bagirrra123 [75]

Answer:

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= (1+ 0.15) x = 1.15x

15% = 0.15 , added to x = 1.15x

4 0
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