Write the ratio explain what the ratio means bottles to cans 5 can 2 bottles

Mathematics

1 answer:

daser333 [38]3 years ago

5 0

For every 5 cans there are 2 bottles

You might be interested in

In which quadrant does the terminal side of a 314° angle lie?

alexandr1967 [171]

We know that

I Quadrant -----> for angles between 0 degrees and 90 degrees
II Quadrant-----> for angles between 90 degrees and 180 degrees
III Quadrant-----> for angles between 180 degrees and 270 degrees
iV Quadrant-----> for angles between 270 degrees and 360 degrees

therefore

314 degrees belong to the IV quadrant

5 0

4 years ago

Cyber Monday Shopping A survey of U.S. adults found that of people said that they would get no work done on Cyber Monday since t

Delicious77 [7]

Complete question :

Cyber Monday Shopping A survey of 1000 U.S. adults found that 33% of people said that they would get no work done on Cyber Monday since they would spend all day shopping online. Find the 95% confidence interval of the true proportion.

Answer:

(0.301 ; 0.359)

Step-by-step explanation:

Sample size, n = 1000

Phat = 0.33

(1 - phat) = 1 - 0.33 = 0.67

Zcritical at 95% = 1.96

Confidence interval :

Phat ± margin of error

Margin of Error = Zcritical * sqrt[(phat(1 - phat)) /n]

Margin of Error = 1.96 * sqrt[(0.33 * 0.67) / 1000]

Margin of Error = 1.96 * 0.0148694

Margin of Error = 0.0291440

Lower boundary = (0.33 - 0.0291440) = 0.300856

Upper boundary = (0.33 + 0.0291440) = 0.359144

(0.301 ; 0.359)

3 0

3 years ago

Suppose C1;C2;C3 are three didifferent biased coins, whose probability of heads equals 0.4, 0.5, and 0.2 respectively. Suppose c

allochka39001 [22]

Solution :

$P(A \cap B) = P(\text{coin}\ C_1 \text{ chosen and 4 heads in 10 trails})$

$= \frac{1}{3} \times ^{10}C_4 \times 0.4^4 \times 0.6^6$

= 0.0836

P(B)= P(coin $C_1$ chosen and 4 heads in 10 trails) + P(coin $C_2$ chosen and 4 heads in 10 trail) + P(coin $C_3$ chosen and 4 heads in 10 trail)

$=\frac{1}{3}\times ^{10}C_4 \times 0.4^4 \times 0.6^6 + \frac{1}{3}\times ^{10}C_4 \times 0.5^4 \times 0.5^6 + \frac{1}{3}\times ^{10}C_4 \times 0.2^4 \times 0.8^6$