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1 year ago

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A local high school held a two-day fundraiser selling lunches. The equations shown represent the amount of money the school coll

ected during the two days. {12x + 4y = 15232x + 12y = 420 In the equations, x represents the cost of a chicken lunch and y represents the cost of a vegetarian lunch. Which is the cost of the vegetarian lunch?

1 answer:

kramer1 year ago

8 0

Answer:

11

Step-by-step explanation:

The two equations appear to be ...

12x +4y = 152
32x +12y = 420

These can be solved for y using Cramer's rule:

y = (152(32) -420(12))/(4(32) -12(12)) = -176/-16 = 11

The cost of the vegetarian lunch is 11.

_____

<em>Comment on Cramer's Rule</em>

For equations ...

ax +by =c

dx +ey = f

The solutions are ...

x = (bf -ey)/(bd -ea)

y = (cd -fa)/(bd -ea) . . . . note the denominators are the same expression

Once you memorize the pattern of products, this can be the simplest way to solve a pair of equations--especially if you only need one of the variable values.

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Find a right triangle that has legs with your irrational Lengths and a hypotenuse with a rational Length

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gg easy

the legs of a triangle are related as $a^2+b^2=c^2$ wehre a, b, and c are the legs and the hypotonuse of a triangle respectively

let's find some stuff

one easy way is to find a rational value for c find values of a and b

lets pick c=5

$a^2+b^2=5^2$

$a^2+b^2=25$

now split up 25 into 2 things that don't have rational square roots

if we say $a^2=19$ and $b^2=6$ then we get

$a=\sqrt{19}$ and $b=\sqrt{6}$ (which are both irrational)

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1 year ago

A friend asks to borrow $900 for three months and agrees to pay 5% simple interest per year. How much interest will you earn in

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Answer:

$11.25 intrest

Step-by-step explanation:

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1 year ago

In 2009, a city with a population of 796,000 had 2,388 births. What is the number of births per 10,000 in the city?

Tom [10]

796,000/2388=10,000/x
796,000x= (2388*10,000)
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2 years ago

Each vertical cross-section of the triangular prism shown below is an isosceles triangle.

vova2212 [387]

Answer:

Height is 3

Step-by-step explanation:

4.24 x 4.24 x 6

Right triangle base = a + b + c

a^2 + 3^2 = 4.24^2

= a^2 + 9 = 17.98

We cross out b to subtract.

a^2 = 17.98 - 9

a^2 = 8.98

We then square √a^2 = √8.98

a = 2.996

We round up a = 3

We have found the height is 3

5 0

2 years ago

During April of 2013, Gallup randomly surveyed 500 adults in the US, and 47% said that they were happy, and without a lot of str

Brilliant_brown [7]

Answer:

number of successes

$k = 235$

number of failure

$y = 265$

The criteria are met

A

The sample proportion is $\r p = 0.47$

B

$E =4.4 \%$

C

What this mean is that for N number of times the survey is carried out that the which sample proportion obtain will differ from the true population proportion will not more than 4.4%

Ci

$r = 0.514 = 51.4 \%$

$v = 0.426 = 42.6 \%$

D

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

E

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

F

Yes our result would support the claim because

$\frac{1}{3 } \ of N < \frac{1}{2} (50\%) \ of \ N , \ Where\ N \ is \ the \ population\ size$

Step-by-step explanation:

From the question we are told that

The sample size is $n = 500$

The sample proportion is $\r p = 0.47$

Generally the number of successes is mathematical represented as

$k = n * \r p$

substituting values

$k = 500 * 0.47$

$k = 235$

Generally the number of failure is mathematical represented as

$y = n * (1 -\r p )$

substituting values

$y = 500 * (1 - 0.47 )$

$y = 265$

for approximate normality for a confidence interval criteria to be satisfied

$np > 5 \ and \ n(1- p ) \ >5$

Given that the above is true for this survey then we can say that the criteria are met

Given that the confidence level is 95% then the level of confidence is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5 \%$

$\alpha =0.05$

Next we obtain the critical value of $\frac{\alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{ \alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as

$E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{\r p (1- \r p}{n} }$

substituting values

$E = 1.96 * \sqrt{ \frac{0.47 (1- 0.47}{500} }$

$E = 0.044$

=> $E =4.4 \%$

What this mean is that for N number of times the survey is carried out that the proportion obtain will differ from the true population proportion of those that are happy by more than 4.4%

The 95% confidence interval is mathematically represented as

$\r p - E < p < \r p + E$

substituting values

$0.47 - 0.044 < p < 0.47 + 0.044$

$0.426 < p < 0.514$

The upper limit of the 95% confidence interval is $r = 0.514 = 51.4 \%$

The lower limit of the 95% confidence interval is $v = 0.426 = 42.6 \%$

This 95% confidence interval mean that the the chance of the true population proportion of those that are happy to be exist within the upper and the lower limit is 95%

Given that 50% of the population proportion lie with the 95% confidence interval the it correct to say that it is reasonably likely that a majority of U.S. adults were happy at that time

Yes our result would support the claim because

$\frac{1}{3 } < \frac{1}{2} (50\%)$

3 0

1 year ago