Parallel lines s and t are cut by a transversal r.

Mathematics

1 answer:

ankoles [38]3 years ago

4 0

Answer:

s and t are equal

Step-by-step explanation:

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Mitchell drew 56 pictures. Rafi drew p fewer pictures than Mitchell. Write an expression that shows how many pictures Rafi drew.

Angelina_Jolie [31]

This would be 56 - p because 'fewer' indicates subtraction.

8 0

3 years ago

14(0.89)^x+4.5 =162

lisabon 2012 [21]

$14(0.89)^{x + 4.5} = 162$
$\frac{14(0.89)^{x + 4.5}}{14} = \frac{162}{14}$
$0.89^{x + 4.5} = 11\frac{4}{7}$
$ln(0.89^{x + 4.5}) = ln(11\frac{4}{7})$
$(x + 4.5)ln(0.89) = ln(11\frac{4}{7})$
$\frac{(x + 4.5)ln(0.89)}{ln(0.89)} = \frac{ln(11\frac{4}{7})}{ln(0.89)}$
$x + 4.5 = -2101.14032500$
$x = 2105.64032500$

3 0

3 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 $N(113, \ 12)$ ,

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