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Phantasy [73]
3 years ago
7

A car travled 200 miles in 4 hours . What was the car average rate of speed in miles per hour ?

Mathematics
2 answers:
statuscvo [17]3 years ago
8 0
The answer would be 50miles/h
Dahasolnce [82]3 years ago
7 0

rate = 200 miles/ 4 hours = 50 miles/hour

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Please help me with this problem <br> Evaluate (-½·3)³
-Dominant- [34]

Answer: -27/8

Step-by-step explanation:

\left(-\frac{1}{2}\cdot \:3\right)^3

\mathrm{Apply\:exponent\:rule}:\quad \left(-a\right)^n=-a^n,\:\quad \mathrm{if\:}n\mathrm{\:is\:odd}

\left(-\frac{1}{2}\cdot \:3\right)^3=-\left(\frac{1}{2}\cdot \:3\right)^3

=-\left(\frac{1}{2}\cdot \:3\right)^3

\mathrm{Apply\:exponent\:rule}:\quad \left(a\cdot \:b\right)^n=a^nb^n

\left(\frac{1}{2}\cdot \:3\right)^3=\left(\frac{1}{2}\right)^3\cdot \:3^3

=-\left(\frac{1}{2}\right)^3\cdot \:3^3

=-\frac{1}{8}\cdot \:3^3

=-\frac{1}{8}\cdot \:27

=-\frac{27}{8}

3 0
2 years ago
Nolan used the following procedure to find an estimate for StartRoot 18 EndRoot.
stiks02 [169]

Answer:

Nolan correctly identified the square  numbers before and after 18.

The square roots of them are 4 and 5.

Clearly, square root of 18 should lie between 4 and 5 only.

He, then carefully squared 4.1, 4.2, 4.3 etc. and identified that 4.3 squared is nearer to 18.

Since, Nolan is finding estimated square root, his steps are cool and he didn't make any error.

4 0
2 years ago
Read 2 more answers
Best known for its testing program, ACT, Inc., also compiles data on a variety of issues in education. In 2004 the company repor
GarryVolchara [31]

Answer:

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

Step-by-step explanation:

Check for conditions

For this case in order to use the normal distribution for this case or the 68-95-99.7% rule we need to satisfy 3 conditions:

a) Independence : we assume that the random sample of 400 students each student is independent from the other.

b) 10% condition: We assume that the sample size on this case 400 is less than 10% of the real population size.

c) np= 400*0.74= 296>10

n(1-p) = 400*(1-0.74)=104>10

So then we have all the conditions satisfied.

Solution to the problem

For this case we know that the distribution for the population proportion is given by:

p \sim N(p, \sqrt{\frac{p(1-p)}{n}})

So then:

\mu_p = 0.74

\sigma_p =\sqrt{\frac{0.74(1-0.74)}{400}}=0.0219

The empirical rule, also referred to as the three-sigma rule or 68-95-99.7 rule, is a statistical rule which states that for a normal distribution, almost all data falls within three standard deviations (denoted by σ) of the mean (denoted by µ). Broken down, the empirical rule shows that 68% falls within the first standard deviation (µ ± σ), 95% within the first two standard deviations (µ ± 2σ), and 99.7% within the first three standard deviations (µ ± 3σ).

\mu_p -\sigma_p = 0.74-0.0219=0.718

\mu_p +\sigma_p = 0.74+0.0219=0.762

68% of the rates are expected to be betwen 0.718 and 0.762

\mu_p -2*\sigma_p = 0.74-2*0.0219=0.696

\mu_p +2*\sigma_p = 0.74+2*0.0219=0.784

95% of the rates are expected to be betwen 0.696 and 0.784

\mu_p -3*\sigma_p = 0.74-3*0.0219=0.674

\mu_p +3*\sigma_p = 0.74+3*0.0219=0.806

99.7% of the rates are expected to be betwen 0.674 and 0.806

5 0
3 years ago
Mr. Jones sent his kids to Mt. Olympus. He gave each of his 5 children money for the day that included money for 4 drinks, 3 san
mihalych1998 [28]

Answer:

195

Step-by-step explanation:

4 drinks = 1.5 * 4 = 6.

3 sandwiches = 2 * 3 = 6

2 treats = 1 * 2 = 2

$6+$6+$2+$25 = 39

Multiply by 5 for each kid

39*5=195

6 0
2 years ago
Anyone who can help me out with the above question​
Citrus2011 [14]

Answer:

Simplified: 0.024

Step-by-step explanation:

5 0
3 years ago
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