The number line shows the distance in meters of two birds ,A and B from a worm located at point x:

Lainey bought a set of 20 makers for $6 what is the cost of 1 marker? <br>

loris [4]

Answer:

0.30

Step-by-step explanation:

$6 Total

20 markers

6/20=0.30

0.3 per maker.

Cost of 1 marker is $0.30

6 0

3 years ago

Divide the difference of 18 and 3 by the sun of 1 and 2

NikAS [45]

Okay.

(18-3)/(1+2) = 15/3 = 5

answer: 5

4 0

3 years ago

Read 2 more answers

Joe can cut and split a cord of firewood in 3 fewer hours than Dwight can. When they work together, it takes them 2 hours. How

aalyn [17]

Answer:

Dwight will take 6 hours to finish the job alone and Joe will take 3 hours to finish the job alone.

Step-by-step explanation:

Let us assume the time taken by Dwight to split a cord of firewood = K hrs

So, the per hour rate of Dwight = $(\frac{1}{K})$

As, Joe uses 3 LESS hours then Dwight.

So, the time taken by Joe to split a cord of firewood = (K- 3) hrs

So, the per hour rate of Joe = $(\frac{1}{K-3})$

Now, when both of them wok together, it takes them 2 hours.

So, the per hour rate of BOTH of them = $(\frac{1}{2})$

⇒ Per hour rate of ( Dwight + Joe) = $(\frac{1}{2} )$

$\implies (\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})$

Now, solving for the value of K , we get:

$(\frac{1}{K}) + (\frac{1}{K-3}) = (\frac{1}{2})\\\implies \frac{(K-3) + K}{K (K-3)} = (\frac{1}{2})\\\implies 2(2K -3) = K^2 - 3K\\\implies k^2 - 3K -4K +6 = 0\\\implies K^2 - 7K + 6= 0\\\implies K^2 - 6K - K + 6= 0\\\implies K(K-6) -1(K - 6)= 0\\\implies (K-6)(K-1) = 0$

Implies either K = 6 Or K = 1

But if K = 1, (K-3) = 1- 3 = -2 hours would be A CONTRADICTION.

⇒ K = 6 hours

Hence, Dwight will take 6 hours to finish the job alone and Joe will take (k-3) = (6-3) = 3 hours to finish the job alone.

8 0

3 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

So

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