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labwork [276]
2 years ago
7

1200 cm per hour= ? Km per week

Mathematics
2 answers:
Bond [772]2 years ago
5 0

Answer: 2.016 km per week

Step-by-step explanation:

Given expression : 1200 cm per hour

We know that 1\text{ km}=100000\text{cm}\\\\\Rightarrow\ 1\text{ cm}=\dfrac{1}{100000}\text{ km}

Also,

1\text{ day}=24\text{ hours} \\\\\Rightarrow\ \text{1 week}=7\times24\text{ hours}=168\text{hours}\\\\\Rightarrow\ \text{1 hour}=\dfrac{1}{68}\text{ week}

\text{ 1200 cm per hour}=\dfrac{1200}{100000}\times\dfrac{1}{\dfrac{1}{168}}\\\\=\dfrac{1200}{100000}\times168=2.016\text{ km per week}

Elis [28]2 years ago
4 0

Answer:

.012km/hr

Step-by-step explanation:

(1200cm)(1m/100cm)(1km/1000m)=0.012km/hr

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A tunnel is in the shape of a parabola. The maximum height is 16 m and it is 16 m wide at the base, as shown below.
NeX [460]
Vertex at the origin and opening down → y=ax^2
Width: w=16
x=w/2→x=16/2→x=8
x=8, y=-16→y=ax^2→-16=a(8)^2→-16=a(64)→-16/64=a(64)/64→-1/4=a→a=-1/4

y=ax^2→y=-(1/4)x^2

7 m from the edge of the tunnel → x=w/2-7=8 m-7 m→x=1 m
x=1→y=-(1/4)x^2=-(1/4)(1)^2=-(1/4)(1)→y=-1/4

Vertical clearance: 16-1/4=16-0.25→Vertical clearance=15.75 m

Please, see the attached file.

Answer: Third option 15.75 m

3 0
2 years ago
Read 2 more answers
Thomas wants to invite Madeline to a party. He has an 80% chance of bumping into her at school. Otherwise, he’ll call her on the
Ivenika [448]

Answer:

84%

Step-by-step explanation:

The probability of Thomas bumping into her at school is 80%, so the probability of not bumping into her is 100% - 80% = 20%.

If he doesn't bump into her (20% chance), he will call her, and the probability of asking her in this case is 60%, so the final probability of asking her in this case is:

P_1 = 20\% * 60\% = 12\%

If he bumps into her (80% chance), the probability of asking her is 90%, so the final probability of asking her in this case is:

P_2 = 80\% * 90\% = 72\%

To find the probability of Thomas inviting Madeline to the party, we just have to sum the probabilities we found above:

P = P_1 + P_2

P = 12\% + 72\% = 84\%

5 0
2 years ago
Find the interest due to the bank on a loan of $1000 at 7.5% for 280 days
Lera25 [3.4K]

Answer:

The interest is \$57.53

Step-by-step explanation:

we know that

The simple interest formula is equal to

I=P(rt)

where

I is the Final Interest Value

P is the Principal amount of money to be invested

r is the rate of interest  

t is Number of Time Periods

in this problem we have

t=280/365\ years\\ P=\$1.000\\r=0.075

substitute in the formula above

I=\$1.000(0.075*280/365)

I=\$57.53

7 0
2 years ago
PLEASE HELP ASAP: A particle is moving with velocity v(t) = t2 – 9t + 18 with distance, s measured in meters, left or right of z
dimaraw [331]
1) \frac{v(8)-v(0)}{8 - 0} = \frac{10-18}{8} = -1\frac{m}{s}.
2) v(5) = 5^2-9*5+18 = 25-45+18 = -2\frac{m}{s}.
3) The particle is moving right when the velocity function is positive: 0\ \textless \ t\ \textless \ 3 or 6\ \textless \ t\ \textless \ 8.
4) When 0\ \textless \ t\ \textless \ \frac{9}{2} the particle is slowing down because the acceleration is close to zero \Rightarrow the particle is speeding up when acceleration is increasing away from zero: \frac{9}{2}\ \textless \ t\ \textless \ 8.
5) (\frac{1}{8})* \int\limits^8_0 {t^2-9t+18dt}=\frac{1}{8}*(\frac{t^3}{3}-(\frac{9}{2}*t^2+18t)_{0}^{8}= \\=(\frac{1}{8})*(\frac{8^3}{3}-(\frac{9}{2})*8^2+18*8)=\frac{8^2}{3}-(\frac{9}{2})*8+18=3\frac{1}{3} \frac{m}{s}.
3 0
2 years ago
Crossett Trucking Company claims that the mean weight of its delivery trucks when they are fully loaded is 6,000 pounds and the
solniwko [45]

Answer:

(5953.52,6046.49)

Step-by-step explanation:

We are given the following in the question:

Mean, \mu = 6,000 pounds

Sample size, n = 40

Alpha, α = 0.05

Standard deviation, σ = 150 pounds

95% Confidence interval:

\mu \pm z_{critical}\frac{\sigma}{\sqrt{n}}

Putting the values, we get,

z_{critical}\text{ at}~\alpha_{0.05} = 1.96

6000 \pm 1.96(\dfrac{150}{\sqrt{40}} )\\\\ = 6000 \pm 46.4854=\\(5953.5146,6046.4854)\approx (5953.52,6046.49)

are the limits within which 95% of the sample means occur.

6 0
2 years ago
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