-3s is greater than or less than 33

kodGreya [7K]

0.7 is bigger.........

7 0

3 years ago

Help asap plz ill give a gold medal. the label on cars antifreeze claims to protect the car between -30celsius and 130celsius. t

irga5000 [103]

The inequality would start out looking like this:
$-30\ \textless \ \frac{5}{9} (F-32)\ \textless \ 130$
Now it's just a matter of solving the inequalities simultaneously. Get rid of the fraction by multiplying everything by 9:
$-270\ \textless \ 5(F-32)\ \textless \ 1170$
Then distribute the 5 into the parenthesis:
$-270\ \textless \ 5F-160\ \textless \ 1170$
Now add 160 everywhere:
$-110\ \textless \ 5F\ \textless \ 1330$
and finally divide everything by 5:
-22<F<266

4 0

3 years ago

There are 6 diagonals that can be drawn from one vertex of an octagon, true or false.

FinnZ [79.3K]

It's False; an octagon has 8 vertices. When you remove the starting vertex and the two adjacent vertices we're left with 5 possible diagonals

6 0

3 years ago

Solve for x. z = 6 π x y z/6πy =x z/6π =x x = 6 π y z z/6 =x

IceJOKER [234]

The answer should be X = Z / 6piy

6 0

3 years ago

One measure of an athlete’s ability is the height of his or her vertical leap. Many professional basketball players are known fo

almond37 [142]

Answer:

(1) P( $\bar X$ < 26 inches) = 0.0436

(2) P(27.5 inches < $\bar X$ < 28.5 inches) = 0.2812

Step-by-step explanation:

We are given that the mean vertical leap of all NBA players is 28 inches. Suppose the standard deviation is 7 inches and 36 NBA players are selected at random.

Firstly, Let $\bar X$ = mean vertical leap for the 36 players

Assuming the data follows normal distribution; so the z score probability distribution for sample mean is given by;

Z = $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean vertical leap = 28 inches

$\sigma$ = standard deviation = 7 inches

n = sample of NBA player = 36

(1) Probability that the mean vertical leap for the 36 players will be less than 26 inches is given by = P( $\bar X$ < 26 inches)

P( $\bar X$ < 26) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{26-28}{\frac{7}{\sqrt{36} } }$ ) = P(Z < -1.71) = 1 - P(Z $\leq$ 1.71)

= 1 - 0.95637 = 0.0436

(2) <em>Now, here sample of NBA players is 26 so n = 26.</em>

Probability that the mean vertical leap for the 26 players will be between 27.5 and 28.5 inches is given by = P(27.5 inches < $\bar X$ < 28.5 inches) = P( $\bar X$ < 28.5 inches) - P( $\bar X$ $\leq$ 27.5 inches)

P( $\bar X$ < 28.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ < $\frac{28.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z < 0.36) = 0.64058 {using z table}

P( $\bar X$ $\leq$ 27.5) = P( $\frac{\bar X - \mu}{\frac{\sigma}{\sqrt{n} } }$ $\leq$ $\frac{27.5-28}{\frac{7}{\sqrt{26} } }$ ) = P(Z $\leq$ -0.36) = 1 - P(Z < 0.36)

= 1 - 0.64058 = 0.35942

Therefore, P(27.5 inches < $\bar X$ < 28.5 inches) = 0.64058 - 0.35942 = 0.2812

6 0

3 years ago