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2 years ago

Is the answer 250 or 500?

Mathematics

2 answers:

kow [346]2 years ago

5 0

Answer:

for what? I am just asking because I don't see the question or anything like that

iren [92.7K]2 years ago

3 0

Answer:

What is the question?

Step-by-step explanation:

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@mehek14

gayaneshka [121]

X=cost per hamburgurs
y=cost per hot dog

200x+150y=1450
200x+250y=1750

we can multiply first equation by -1 and add to 2nd

-200x-150y=-1450
200x+250y=1750 +
0x+100y=300

100y=300
divide both sides by 100
y=3

sub back

200x+150y=1450
200x+150(3)=1450
200x+450=1450
minus 450 both sides
200x=1000
divide both sides by 200
x=5

the cost of a hamburgur is $5
the cost of a hot dog is $3

8 0

3 years ago

Read 2 more answers

Rational exponents: product rule simplify v^3/4 multiply v^5/7

marusya05 [52]

Answer:

$v\sqrt[28]{v^{13}}$

Step-by-step explanation:

The exponent rules that apply are ...

$(a^b)(a^c)=a^{b+c}\\\\a^{\frac{b}{c}}=\sqrt[c]{a^b}$

Using these rules for your product, we have ...

$v^{\frac{3}{4}}\times v^{\frac{5}{7}}=v^{\frac{3}{4}+\frac{5}{7}}\\\\=v^{\frac{41}{28}}=v\cdot v^{\frac{13}{28}}=v\sqrt[28]{v^{13}}$

6 0

3 years ago

Which points lie in the second quadrant? Check all that apply.

Tomtit [17]

Hello there! Point E and Point F are in the second quadrant.

The second quadrant is the first quadrant you see (upper right, see image attached for all quadrants). In this quadrant, you can see the points E and F. These are your correct answers.

I hope this helps and have a great day!

4 0

3 years ago

Read 2 more answers

Discrete R.V. Assume a student shows up for EGR 280 completely unprepared and the instructor gives a pop quiz. Assume there are

Tom [10]

Answer:

a) p=0.2

b) probability of passing is 0.01696

c) The expected value of correct questions is 1.2

Step-by-step explanation:

a) Since each question has 5 options, all of them equally likely, and only one correct answer, then the probability of having a correct answer is 1/5 = 0.2.

b) Let X be the number of correct answers. We will model this situation by considering X as a binomial random variable with a success probability of p=0.2 and having n=6 samples. We have the following for k=0,1,2,3,4,5,6

$P(X=k) = \binom{n}{k}p^{k}(1-p)^{n-k} = \binom{6}{k}0.2^{k}(0.8)^{6-k}$ .

Recall that $\binom{n}{k}= \frac{n!}{k!(n-k)!}$ In this case, the student passes if X is at least four correct questions, then

$P(X\geq 4) = P(X=4)+P(X=5)+P(X=6)=\binom{6}{4}0.2^{4}(0.8)^{6-4}+\binom{6}{5}0.2^{5}(0.8)^{6-5}+\binom{6}{6}0.2^{6}(0.8)^{6-6}= 0.01696$

c)The expected value of a binomial random variable with parameters n and p is $E[X] = np$ . IN our case, n=6 and p =0.2. Then the expected value of correct answers is $6\cdot 0.2 = 1.2$