answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Helga [31]
2 years ago
9

Amir drove from Jerusalem down to the lowest place on Earth, the Dead Sea, descending at a rate of 12 meters per minute. He was

at sea level after 30 minutes of driving. Graph the relationship between Amir's altitude relative to sea level (in meters) and time (in minutes).

Mathematics
2 answers:
SSSSS [86.1K]2 years ago
6 0
<h2>Answer:</h2>

First of all let's write the slope-intercept form of the equation of a line, which is:

y=mx+b \\ \\ Where: \\ \\ m: \ slope \\ \\ b: \ y-intercept

So we just need to find m \ and \ b to solve this problem.

Moreover, this problem tells us that Amir drove from Jerusalem down to the lowest place on Earth, the Dead Sea, descending at a rate of 12 meters per minute. So this rate is the slope of the line, that is:

m=-12

Negative slope because Amir is descending. So:

y=-12x+b

To find b, we need to use the information that tells us that he was at sea level after 30 minutes of driving, so this can be written as the point (30,0). Therefore, substituting this point into our equation:

y=-12x+b \\ \\ 0=-12(30)+b \\ \\ 0=-360+b \\ \\ \therefore b=360

Finally, the equation of Amir's altitude relative to sea level (in meters) and time (in minutes) is:

\boxed{y=-12x+360}

Whose graph is shown bellow.

pantera1 [17]2 years ago
4 0

Answer:

-12+360

Step-by-step explanation:

You might be interested in
Rocky Mountain National Park is a popular park for outdoor recreation activities in Colorado. According to U.S. National Park Se
Ugo [173]

Answer:

a) 0.6628 = 66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance

b) 0.5141 = 51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

c) 0.5596 = 55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

d) 0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 175

(a) What is the probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance?

46.7% of visitors to Rocky Mountain National Park in 2018 entered through the Beaver Meadows. This means that p = 0.467. So

\mu = E(X) = np = 175*0.467 = 81.725

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.467*0.533} = 6.6

This probability, using continuity correction, is P(X \geq 85 - 0.5) = P(X \geq 84.5), which is 1 subtracted by the pvalue of Z when X = 84.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{84.5 - 81.725}{6.6}

Z = 0.42

Z = 0.42 has a pvalue of 0.6628.

66.28% probability that at least 85 visitors had a recorded entry through the Beaver Meadows park entrance.

(b) What is the probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance?

Using continuity correction, this is P(80 - 0.5 \leq X <  90 - 0.5) = P(79.5 \leq X \leq 89.5), which is the pvalue of Z when X = 89.5 subtracted by the pvalue of Z when X = 79.5. So

X = 89.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{89.5 - 81.725}{6.6}

Z = 1.18

Z = 1.18 has a pvalue of 0.8810.

X = 79.5

Z = \frac{X - \mu}{\sigma}

Z = \frac{79.5 - 81.725}{6.6}

Z = -0.34

Z = -0.34 has a pvalue of 0.3669.

0.8810 - 0.3669 = 0.5141

51.41% probability that at least 80 but less than 90 visitors had a recorded entry through the Beaver Meadows park entrance

(c) What is the probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance?

6.3% of visitors entered through the Grand Lake park entrance, which means that p = 0.063

\mu = E(X) = np = 175*0.063 = 11.025

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.063*0.937} = 3.2141

This probability, using continuity correction, is P(X < 12 - 0.5) = P(X < 11.5), which is the pvalue of Z when X = 11.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{11.5 - 11.025}{3.2141}

Z = 0.15

Z = 0.15 has a pvalue of 0.5596.

55.96% probability that fewer than 12 visitors had a recorded entry through the Grand Lake park entrance.

(d) What is the probability that more than 55 visitors have no recorded point of entry?

22.7% of visitors had no recorded point of entry to the park. This means that p = 0.227

\mu = E(X) = np = 175*0.227 = 39.725

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{175*0.227*0.773} = 5.54

Using continuity correction, this probability is P(X \leq 55 + 0.5) = P(X \leq 55.5), which is the pvalue of Z when X = 55.5. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{55.5 - 39.725}{5.54}

Z = 2.85

Z = 2.85 has a pvalue of 0.9978

0.9978 = 99.78% probability that more than 55 visitors have no recorded point of entry

8 0
2 years ago
Which of the following shows the extraneous solution to the logarithmic equation below???
avanturin [10]
Short Answer x = - 6
Remark
You gave the second best answer. 

The trick here is not to divide both sides by 2. Solve the problem this way.

log_5(x + 1)^2 = 2 Take the antilog of both sides
(x + 1)^2 = 5^2       Expand the equation
x^2 + 2x + 1 = 25   Subtract 25 from both sides.
x^2 + 2x - 24 = 0    Factor
(x + 6)(x - 4) = 0     Find the zeros.

x + 6 = 0
x = - 6  <<<<<<<< Answer. This is the extraneous root.
The reason this is an extraneous root is that x<=0 do not have a logarithem

x - 4 = 0
x = 4    This is a legitimate result to the original equation.
7 0
2 years ago
Read 2 more answers
Find the ratio of $22 to $5.50
solniwko [45]

Answer:

4:1

Step-by-step explanation:

22/5.50

divide numerator by denominator and denominator by denominator

22/5.50=4

5.50/5.50=1

4/1 or 4:1

4 0
2 years ago
A copper smelting process is supposed to reduce the arsenic content of the copper to less than 1000 ppm. let μ denote the mean a
Hitman42 [59]
The rejection region is give by 

|z_{test}|\ \textgreater \ z_{\alpha/2}

where the test statistics is given by

\frac{\bar{x}-\mu}{\sigma/\sqrt{n}} = \frac{980-1000}{100/\sqrt{75}} \\ \\ = \frac{-20}{100/8.6603} = \frac{-20}{11.5470} =-1.73

i.e. |z_{test}|=|-1.73|=1.73

Thus, z_{\alpha/2}=1.73

Using the statistical table, the level of the test is 0.04.
6 0
2 years ago
Two sandboxes with the same area are shown. The equation w(3w+1)=5^2 represents the area of Sandbox 2 in terms of its width. Whi
liubo4ka [24]

sandbox 1 area = 5*5 = 25 square m

sandbox 2

25=w*(3w+1)=

3w^2+w-25=0

w=2.7248 (round to 2.72 m)

2.72*3=8.16+1 = 9.16

Longest side = 9.16 meters

6 0
2 years ago
Read 2 more answers
Other questions:
  • a school conducts 27 test in 36 weeks. assume the school conducts tests at a constant rate .what is the slope of the line that r
    8·2 answers
  • .1.In a large bag of marbles, 25% of them are red. A child chooses 4 marbles from this bag. If the child chooses the marbles at
    10·2 answers
  • Suppose you buy a CD for $300 that earns 3% APR and is compounded quarterly. The CD matures in 3 years. How much will this CD be
    14·2 answers
  • The weight, w, of a spring in pounds is given by 0.9 times the square root of the energy, E, stored by the spring in joules. If
    15·2 answers
  • The formula uppercase S = StartFraction n (a Subscript 1 Baseline plus a Subscript n Baseline) Over 2 EndFraction gives the part
    15·2 answers
  • Determine all real numbers a$ such that the inequality $ |x^2 2ax 3a|\le2$ has exactly one solution in x$.
    5·1 answer
  • One of the giraffes, Bernard, is sick. Last week he weighed 2,490 pounds. This week he weighs 2,312 pounds. If w represents the
    14·2 answers
  • a bricklayer needs to order 6300kg of building sand.one grain of this sand weighs 7x10^-5. how many grains of sand are there in
    14·1 answer
  • Scores on Ms. Bond's test have a mean of 70 and a standard deviation of 11. David has a score of 52 on Ms. Bond's test. Scores o
    5·1 answer
  • Cameron has a mask collection. He keeps 267 of the masks on his wall, which is 89%
    5·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!