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2 years ago

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Jeremy rides a Ferris wheel. The graph shows h, Jeremy’s height above the ground at any time during the ride. Which inequality r

epresents the heights shown on the graph? 10 < h < 50 10 ≤ h ≤ 50 h > 10 or h < 50 h ≥ 10 or h ≤ 50

2 answers:

djverab [1.8K]2 years ago

8 0

Answer:

The answer is B, just took the test on edge.

Step-by-step explanation:

chubhunter [2.5K]2 years ago

3 0

I think it is 10<h<50

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A car uses 3t gallons of gasoline to travel 5 miles.

Nadusha1986 [10]

Answer: a) $\frac{24}{5}t gallons$ are needed to travel 8 miles

b) Town X and town Y are 15 miles apart.

Step-by-step explanation:

Given : Gasoline used to travel 5 miles = 3t gallons

a) To find: Gasoline used to travel 8 miles = ?

Using unitary method:

If Gasoline used to travel 5 miles = 3t gallons

Thus Gasoline used to travel 8 miles = $\frac{3t}{5}\times 8=\frac{24}{5}t gallons$

$\frac{24}{5}t gallons$ are needed to travel 8 miles

b) when 3t gallons of gasoline are used , distance covered = 5 miles

Thus when 9t gallons of gasoline are used , distance covered = $\frac{5}{3t}\times 9t=15 miles$

Thus town X and town Y are 15 miles apart.

6 0

2 years ago

The areas of the squares created by the side lengths of the triangle are shown. Which best explains whether this triangle is a r

nikitadnepr [17]

Answer:

Based on the converse of the Pythagorean Theorem, the triangle is not a right triangle, because $9+12\neq 15$

Step-by-step explanation:

The complete question in the attached figure

we know that

If the length sides of a triangle, satisfy the Pythagorean Theorem, then is a right triangle

$c^2=a^2+b^2$

where

c is the hypotenuse (the greater side)

a and b are the legs

In this problem

The length sides squared of the triangle are equal to the areas of the squares

so

$c^2=15\ in^2$

$a^2=12\ in^2$

$b^2=9\ in^2$

substitute

$15=12+9$

$15=21$ ----> is not true

so

The length sides not satisfy the Pythagorean Theorem

therefore

Based on the converse of the Pythagorean Theorem, the triangle is not a right triangle, because $9+12\neq 15$

8 0

2 years ago

Read 2 more answers

Morgan is walking her dog on an 8-meter-long leash. She is currently 500 meters from her house, so the maximum and minimum dista

lara31 [8.8K]

<u>The minimum distance is 492 meters from the house (500 - 8 = 492), and the maximum distance is 508 meters from the house (500 + 8 = 508). The dog may be slightly closer to the house, depending on how long the dog is, or if Morgan is using a leash extender.</u>

3 0

2 years ago

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Josh travels frequently. For a particular airline, it takes 20 minutes for the first bag to arrive in baggage claim after a flig

Orlov [11]

Using the uniform distribution, it is found that:

A,B) 0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

C) 20% of the time does Josh’s bag arrive in less than 22 minutes.

--------------------------

An uniform distribution has two bounds, a and b.

The probability of finding a value of at lower than x is:

$P(X < x) = \frac{x - a}{b - a}$

The probability of finding a value between c and d is:

$P(c \leq X \leq d) = \frac{d - c}{b - a}$

The probability of finding a value above x is:

$P(X > x) = \frac{b - x}{b - a}$

--------------------------

In the graph, we have that the distribution is uniform between 20 and 30 minutes, thus $a = 20, b = 30$

--------------------------

Itens a and b:

Above 27 minutes, thus:

$P(X > 27) = \frac{30 - 27}{30 - 20} = 0.3$

0.3 of the time does it take longer than 27 minutes for Josh’s bag to arrive in baggage claim.

--------------------------

Item c:

Less than 22 minutes, thus:

$P(X < 2) = \frac{22 - 20}{30 - 20} = 0.2$

0.2*100 = 20%

20% of the time does Josh’s bag arrive in less than 22 minutes.

A similar problem is given at brainly.com/question/15855314

6 0

1 year ago

A wheat farmer cuts down the stalks of wheat and gathers them in 200 piles. The 200 gathered piles will be put on a truck. The t

denpristay [2]

Answer:

Check the explanation

Step-by-step explanation:

We want to estimate the total weight of grain on the field based on the data on a simple random sample of 5 piles out of 200. The population and sample sizes are N=200 & n=5 respectively.

1) Let Y_1,Y_2,...,Y_{200} be the weight of grain in the 200 piles and y_1,y_2,...,y_{5} be the weights of grain in the pile from the simple random sample.

We know, the sample mean is an unbiased estimator of the population mean. Therefore,

$\widehat{\mu}=\overline{y}=\frac{1}{5}\sum_{i=1}^{5}y_i=\frac{1}{5}(3.3+4.1+4.7+5.9+4.5)=4.5$

where \mu is the mean weight of grain for all the 200 piles.

Hence, the total grain weight of the population is

$\widehat{Y}=Y_1+Y_2+...+Y_{200}$

= $200\times \widehat{\mu}\: \: \: =200\times 4.5\: \: \: =900\, lbs$

2) To calculate a bound on the error of estimates, we need to find the sample standard deviation.

The sample standard deviation is

S= $\sqrt{\frac{1}{5-1}\sum_{i=1}^{5}(y_i-\overline{y})^2}\: \: \: =0.9486$

Then, the standard error of $\widehat{Y} is$

$\sigma_{\widehat{Y}} =\sqrt{\frac{N^2S^2}{n}\bigg(\frac{N-n}{N}\bigg)}\: \: \:$ =83.785

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{\widehat{Y}}]\: \: \: =[\pm 1.96\times 83.875]\: \: \: =[\pm 164.395]$

3) Let x_1,x_2,...,x_5 denotes the total weight of the sampled piles.

Mean total weight of the sampled piles is

$\overline{x}=\frac{1}{5}\sum_{i=1}^{5}x_i=45$

The sample ratio is

$r=\frac{\overline{y}}{\overline{x}}=\frac{4.5}{45}$ =0.1 , this is also the estimate of the population ratio R= $\frac{\overline{Y}}{\overline{X}} .$

Therefore, the estimated total weight of grain in the population using ratio estimator is

$\widehat{Y}_R\: \: =r\times 8800\: \: =0.1\times 8800\: \: =880\,$ lbs

4) The variance of the ratio estimator is

var(r)= $\frac{N-n}{N}\frac{1}{n}\frac{1}{\mu_x^2}\frac{\sum_{i=1}^{5}(y_i-rx_i)^2}{n-1} , where \mu_x=8800/200=44lbs$

= $\frac{200-5}{200}\, \frac{1}{5}\: \frac{1}{44^2}\, \frac{0.2}{5-1}=0.000005$

Hence, the standard error of the estimate of the total population is

$\sigma_R=\sqrt{X^2 \: var(r)}\: \: \: =\sqrt{8800^2\times 0.000005}\: \: \:$ =21.556

Hence, a 95% bound on the error of estimates is

$[\pm z_{0.025}\times \sigma_{R}]\: \: \: =[\pm 1.96\times 21.556]\: \: \: =[\pm 42.25]$

8 0

2 years ago

Read 2 more answers