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2 years ago

Time (hh:mm) Please enter the displayed to prove I am not a robot :0

Mathematics

1 answer:

masha68 [24]2 years ago

5 0

Answer:

07:45 or 7:45

Step-by-step explanation:

On a clock, the shorter hand shows the hour of the day, while the longer hand shows the minute. The clock is divided into 12 hours and 60 minutes. In this case, the shorter hand is between 7 and 8, and the minute hand is right at 45, meaning that the time is 07:45. Hope this helps!

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A stone is thrown vertically upward from a platform that is 20 feet height at a rate of 160 ft/sec. Use the quadratic function h

Nutka1998 [239]

Check the picture below.

$\textit{vertex of a vertical parabola, using coefficients} \\\\ y=\stackrel{\stackrel{a}{\downarrow }}{-16}x^2\stackrel{\stackrel{b}{\downarrow }}{+160}x\stackrel{\stackrel{c}{\downarrow }}{+20} \qquad \qquad \left(-\cfrac{ b}{2 a}~~~~ ,~~~~ c-\cfrac{ b^2}{4 a}\right)$

$\left(-\cfrac{ 160}{2(-16)}~~~~ ,~~~~ 20-\cfrac{ (160)^2}{4(-16)}\right) \implies \left( - \cfrac{ 160 }{ -32 }~~,~~20 - \cfrac{ 25600 }{ -64 } \right) \\\\\\ \left( 5 ~~~~ ,~~~~ 20 +400 \right)\implies (\stackrel{seconds}{5}~~,~~\stackrel{feet}{420})$

6 0

1 year ago

Tan^2 70° -sec^2 70°

kipiarov [429]

Answer:

-1

Step-by-step explanation:

We need to find the value of tan² 70° -sec² 70°.

We know that,

$\sec^2\theta-\tan^2\theta=1$ ...(1)

We can write the given expression as :

tan² 70° -sec² 70° = -(-tan² 70° +sec² 70°)

=-(sec² 70°-tan² 70°)

Using identity (1)

=-(1)

= -1

Hence, the value of the given expression is -1.

4 0

2 years ago

What is the explicit equation for f(0)=8; f(×)=f(×-1)*3

andriy [413]

Hello from MrBillDoesMath!

Answer:

f(x) = 8 * 3^x

Discussion:

x = 0: f(x) = 8

x = 1: f(1) = f(0) * 3 = 8*3

x = 2: f(2) = f(2-1)*3 = f(1) * 3 = (8*3)*3 = 8 * 3^2

x=3 : f(3) = f(3-1)*3 = f(2)*3 = (8 * 3^2) * 3 = 8 * 3^3

Based on this we guess that

f(x) = 8 * 3^x

Thank you,

MrB

4 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.