12+21 divided by 3 + (2*2 x 0)

Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18Jynessa wants to order these fractions: StartFraction 4 over 9 EndFraction, two-thirds, one-sixth, Negative 2 and one-half. What should she use as her common denominator? 6 9 12 18

Step-by-step explanation:

8 0

2 years ago

Read 2 more answers

Please help me with this :)))

Yuki888 [10]

Answer:

C, 33 1/3%

Step-by-step explanation:

Because there are only two even number that follow this rule: 2<x≥6, and since there are only 6 possible outcomes, the probabilty is 2/6, which is 1/3. In a percent form, this is 100%*2/3, or 33 1/3%.

8 0

2 years ago

The prices of all college textbooks follow a bell-shaped distribution with a mean of $113 and a standard deviation of $12. Using

Soloha48 [4]

Step-by-step explanation:

In statistics, the empirical rule states that for a normally distributed random variable,

68.27% of the data lies within one standard deviation of the mean.

95.45% of the data lies within two standard deviations of the mean.

99.73% of the data lies within three standard deviations of the mean.

In mathematical notation, as shown in the figure below (for a standard normal distribution), the empirical rule is described as

$\Phi(\mu \ - \ \sigma \ \leq X \ \leq \mu \ + \ \sigma) \ = \ 0.6827 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi(\mu \ - \ 2\sigma \ \leq X \ \leq \mu \ + \ 2\sigma) \ = \ 0.9545 \qquad (4 \ \text{s.f.}) \\ \\ \\ \Phi}(\mu \ - \ 3\sigma \ \leq X \ \leq \mu \ + \ 3\sigma) \ = \ 0.9973 \qquad (4 \ \text{s.f.})$

where the symbol $\Phi$ (the uppercase greek alphabet phi) is the cumulative density function of the normal distribution, $\mu$ is the mean and $\sigma$ is the standard deviation of the normal distribution defined as $N(\mu, \ \sigma)$ .

According to the empirical rule stated above, the interval that contains the prices of 99.7% of college textbooks for a normal distribution $N(113, \ 12)$ ,

$\Phi(113 \ - \ 3 \ \times \ 12 \ \leq \ X \ \leq \ 113 \ + \ 3 \ \times \ 12) \ = \ 0.9973 \\ \\ \\ \-\hspace{1.75cm} \Phi(113 \ - \ 36 \ \leq \ X \ \leq \ 113 \ + \ 36) \ = \ 0.9973 \\ \\ \\ \-\hspace{3.95cm} \Phi(77 \ \leq \ X \ \leq \ 149) \ = \ 0.9973$

Therefore, the price of 99.7% of college textbooks falls inclusively between $77 and $149.

5 0

1 year ago