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2 years ago

Find the quotient of 2.632 x 10^4 and 2 x 10^-7

1 answer:

6 0

$(2.632\cdot10^4):(2\cdot10^{-7})=(2.632:2)\cdot(10^4:10^{-7})\\\\=1.316\cdot10^{4-(-7)}=1.316\cdot10^{4+7}=\boxed{1.316\cdot10^{11}}$

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Adam currently runs about 40 miles per week, and he wants to increase his weekly mileage by 30%. How many miles will Adam run pe

BigorU [14]

Answer:

52 miles/week (12 more miles per week)

Step-by-step explanation:

In order to know the miles per week with this increasement, we'll use the rule of 3, assuming that 40 miles is the 100% innitially so:

If:

40 miles --------->100%

X miles ----------> 30%

Solving for X we have the following:

X = 30 * 40 / 100 = 12 miles

So, if Adam wants to increase his weekly mileage by 30%, he needs to run 12 more miles, and that makes a total of 52 miles/week

6 0

2 years ago

What’s a line equation passing through (14,-8) and (24,-3)

Nataly_w [17]

$\bf (\stackrel{x_1}{14}~,~\stackrel{y_1}{-8})\qquad (\stackrel{x_2}{24}~,~\stackrel{y_2}{-3}) \\\\\\ \stackrel{slope}{m}\implies \cfrac{\stackrel{rise} {\stackrel{y_2}{-3}-\stackrel{y1}{(-8)}}}{\underset{run} {\underset{x_2}{24}-\underset{x_1}{14}}}\implies \cfrac{-3+8}{10}\implies \cfrac{5}{10}\implies \cfrac{1}{2}$

$\bf \begin{array}{|c|ll} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-\stackrel{y_1}{(-8)}=\stackrel{m}{\cfrac{1}{2}}(x-\stackrel{x_1}{14}) \\\\\\ y+8=\cfrac{1}{2}x-7\implies y=\cfrac{1}{2}x-15$

4 0

3 years ago

A painter leans a 15 ft ladder against a building. The base of the ladder is 6 ft from the building. To the nearest foot, how hi

stiv31 [10]

Using the Pythagoras theorem

15^2 = x^2 + h^2 where h = height of ladder on the nuiding

h^2 = 15^2 - 6^2 = 189

= 13.75 ft to nearest hundredth

3 0

2 years ago

Read 2 more answers

1. Last week Bradley worked 32 hours

LuckyWell [14K]

Answer:

B ($9.78)

Step-by-step explanation:

5 0

2 years ago

Genetic Defects Data indicate that a particular genetic defect occurs in of every children. The records of a medical clinic show

Dovator [93]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The probability that there exist 60 or more defected children is $P(x \ge 60)=0.0901$

Looking at the value for this probability we see that it is not so small to the point that the observation of this kind would be a rare occurrence

Step-by-step explanation:

From the question we are told that

in every 1000 children a particular genetic defect occurs to 1

The number of sample selected is $n= 50,000$

The probability of observing the defect is mathematically evaluated as

$p = \frac{1}{1000}$

$= 0.001$

The probability of not observing the defect is mathematically evaluated as

$q = 1-p$

$= 1-0.001$

$= 0.999$

The mean of this probability is mathematically represented as

$\mu = np$

Substituting values

$\mu = 50000*0.001$

$= 50$

The standard deviation of this probability is mathematically represented as

$\sigma = \sqrt{npq}$

Substituting values

$= \sqrt{50000 * 0.001 * 0.999}$

$= \sqrt{49.95}$

$= 7.07$

the probability of detecting $x \ge60$ defects can be represented in as normal distribution like

$P(x \ge 60)$

in standardizing the normal distribution the normal area used to approximate $P(x \ge 60)$ is the right of 59.5 instead of 60 because x= 60 is part of the observation

The z -score is obtained mathematically as

$z = \frac{x-\mu }{\sigma }$

$= \frac{59.5 - 50 }{7.07}$

$=1.34$

The area to the left of z = 1.35 on the standardized normal distribution curve is 0.9099 obtained from the z-table shown z value to the left of the standardized normal curve

Note: We are looking for the area to the right i.e 60 or more

The total area under the curve is 1

$P(x \ge 60) \approx P(z > 1.34)$

$= 1-P(z \le 1.34)$

$=1-0.9099$

$=0.0901$

3 0

2 years ago