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2 years ago

A 32 oz. fruit juice label reads: Contains 25% real fruit juice. How many ounces are real fruit juice? 8

Mathematics

2 answers:

olasank [31]2 years ago

8 0

8 ounces, it said it at the end of the question.

AveGali [126]2 years ago

3 0

Yes the answer is 8
if you multiply 32 times 0.25(the percent in decimal form) you get 8

hope this helps :)

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A tune up automotive facility marks up its part by 25% suppose that the facility charges its customers $1.75 for each spark plug

ss7ja [257]

Answer:

$1.4

Step-by-step explanation:

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2 years ago

A blueberry pie is shot out of a pie cannon from a height of 6m and with a velocity of 60 m/s. How many seconds will pass before

sveta [45]

Answer: 1.106 s

Step-by-step explanation:

This situation is related to projectile motion and one the equation that models the height of the blueberry pie in time is:

$y=y_{o}+V_{o}sin(\theta) t-\frac{1}{2}gt^{2}$

Where:

$y=0 m$ is the blueberry pie final height (when it hits the ground)

$y_{o}=6 m$ is the blueberry pie initial height

$V_{o}=60 m/s$ is the blueberry pie initial velocity

$\theta=0 \°$ is the angle, assuming the pie was shot horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

Rewriting the equation:

$0=y_{o}-\frac{1}{2}gt^{2}$

Isolating $t$ :

$t=\sqrt{\frac{2y_{o}}{g}}$

$t=\sqrt{\frac{2(6 m)}{9.8 m/s^{2}}}$

Finally:

$t=1.106 s$

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2 years ago

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Olin [163]

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2 years ago

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Given any two events, E1 and E2, what does the probability P(E1 ∪ E2) represent?

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2 years ago

Suppose a simple random sample of size nequals81 is obtained from a population with mu equals 79 and sigma equals 18. (a) Descr

Daniel [21]

Answer:

(a) The sampling distribution of $\overline{X}$ = Population mean = 79

(b) P ( $\overline{X}$ greater than 81.2 ) = 0.1357

(c) P ( $\overline{X}$ less than or equals 74.4 ) = .0107

(d) P (77.6 less than $\overline{X}$ less than 83.2 ) = .7401

Step-by-step explanation:

Given -

Sample size ( n ) = 81

Population mean $(\nu)$ = 79

Standard deviation $(\sigma )$ = 18

(a) Describe the sampling distribution of $\overline{X}$

For large sample using central limit theorem

the sampling distribution of $\overline{X}$ = Population mean = 79

(b) What is Upper P ( $\overline{X}$ greater than 81.2 ) =

$P(\overline{X}> 81.2)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}> \frac{81.2 - 79}{\frac{18}{\sqrt{81}}})$

= $P(Z> 1.1)$

= $1 - P(Z< 1.1)$

= 1 - .8643 =

= 0.1357

$P(\overline{X}\leq 74.4)$ = $P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{74.4- 79}{\frac{18}{\sqrt{81}}})$

= $P(Z\leq -2.3)$

= .0107

(d) What is Upper P (77.6 less than $\overline{X}$ less than 83.2 ) =

$P(77.6< \overline{X}< 83.2)$ = $P(\frac{77.6- 79}{\frac{18}{\sqrt{81}}})< P(\frac{\overline{X} - \nu }{\frac{\sigma }{\sqrt{n}}}\leq \frac{83.2- 79}{\frac{18}{\sqrt{81}}})$

= $P(- 0.7< Z< 2.1)$

= $(Z< 2.1) - (Z< -0.7)$

= 0.9821 - .2420

= 0.7401

3 0

2 years ago