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lara [203]
2 years ago
5

Round your answer to the nearest hundredth.

Mathematics
1 answer:
BartSMP [9]2 years ago
3 0

Answer:

in \: right \: angle \: triangl \: a \: b \: c \\  tan35 =  \frac{bc}{ac }  =  \frac{bc}{2}  \\ bc = 2 \times 0.7002 \\ bc = 1.4004

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Find the 22nd term. 3, 7.5, 12, 16.5
yaroslaw [1]

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Step-by-step explanation:

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2 years ago
Please Explain how you got your answer!
marysya [2.9K]

Answer:

  19 miles

Step-by-step explanation:

Let m represent the number of miles traveled. Then the delivery charge is ...

  charge = $5.00 + $0.15·m

You want to find the value of m that makes the charge by $7.85, so you want the solution to ...

  $7.85 = $5.00 +$0.15·m

  $2.85 = $0.15·m . . . . . . . . . subtract $5 (this gives the mileage charge)

  $2.85/$0.15 = m = 19 . . . . . divide by the coefficient of m

The number of miles traveled is 19.

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2 years ago
A crew is made up of 8 men; the rest are women. If 32 % of the crew are men, how many people are in the crew?
iVinArrow [24]

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There are 33 people in the crew

5 0
2 years ago
Please help ASAP, giving lots of points.
BARSIC [14]
The answers is the last one

6 0
3 years ago
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
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