All categories

3 years ago

Need help with this question please

Mathematics

2 answers:

irina [24]3 years ago

7 0

You would do KCF so the answer would be 24/6 or 4

PilotLPTM [1.2K]3 years ago

4 0

Answer:

9/16 using KFC

Step-by-step explanation:

if need a better explanation just ask

You might be interested in

A manufacturer of matches randomly and independently puts 23 matches in each box of matches produced. The company knows that one

d1i1m1o1n [39]

Answer:

0.9855 or 98.55%.

Step-by-step explanation:

The probability of each individual match being flawed is p = 0.008. The probability that a matchbox will have one or fewer matches with a flaw is the same as the probability of a matchbox having exactly one or exactly zero matches with a flaw:

$P(X\leq 1)=P(X=0)+P(X=1)\\P(X\leq 1)=(1-p)^{23}+23*(1-p)^{23-1}*p\\P(X\leq 1)=(1-0.008)^{23}+23*(1-0.008)^{23-1}*0.008\\P(X\leq 1)=0.8313+0.1542\\P(X\leq 1)=0.9855$

The probability that a matchbox will have one or fewer matches with a flaw is 0.9855 or 98.55%.

8 0

3 years ago

3. Brenda's school is selling tickets to a spring musical. On the first day of

Natalija [7]

Answer:

The Answer is that Senior Citizen Tickets cost: $4 and Child tickets cost: $7.

Step-by-step explanation:

Let s = the cost of senior citizen tickets

Let c = the cost of child tickets

The number of tickets sold for each type added together equals the sales for each day. Equations below:

Day 1

3s + 9c = $75

Solve for s:

3s = 75 - 9c

s = 25 - 3c

Day 2

8s + 5c = $67

By substitution:

8(25 - 3c) + 5c = 67

200 - 24c + 5c = 67

-19c = -133

c = -133 / -19 = $7 cost for child tickets.

Solve for s:

s = 25 - 3c

s = 25 - 3(7)

s = 25 - 21 = $4 cost for senior citizen tickets.

Proofs:

Day 1

3s + 9c = $75

3(4) + 9(7) = 75

12 + 63 = 75

75 = 75

Day 2

8s + 5c = $67

8(4) + 5(7) = 67

32 + 35 = 67

67 = 67

4 0

2 years ago

What is the following product? 3 root 5 times root 2

Crazy boy [7]

$\bf \textit{rational exponents} \\\\ a^{\frac{ n}{ m}} \implies \sqrt[ m]{a^ n} ~\hspace{10em} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \sqrt[3]{5}\cdot \sqrt{2}\implies \sqrt[3]{5^1}\cdot \sqrt[2]{2^1}\implies 5^{\frac{1}{3}}\cdot 2^{\frac{1}{2}}\implies 5^{\frac{2}{6}}\cdot 2^{\frac{3}{6}}\implies \sqrt[6]{5^2}\cdot \sqrt[6]{2^3} \\\\\\ \sqrt[6]{25}\cdot \sqrt[6]{8}\implies \sqrt[6]{25\cdot 8}\implies \sqrt[6]{200}$