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alex41 [277]
3 years ago
15

Austin is riding his dirt bike at a speed of 350 feet per minute. How many yards would he have traveled in 2.5 hours? (HINT: 3 f

eet in 1 yard)
Mathematics
1 answer:
Harrizon [31]3 years ago
5 0

Answer:

17,500 yards

Step-by-step explanation:

Speed of Austin's bike = 350 ft per minute

Given 3 feet in 1 yard

Now converting speed from  ft/min to yard per minute

3 feet equal 1 yard  (dividing both side by 3)

1 feet equals 1/3 yards

350 ft equals 350*1/3 yards

350 ft equals 350/3 yards

speed of bike in yards per minute = 350/3 yards per minute

______________________________________

Time for which bike traveled = 2.5 hour

one hour has 60 minutes

2.5 hour has 60*2.5 minutes

2.5 hour has 150 minutes

Time for which bike traveled in minutes = 150 minutes

________________________________________

formula for distance = speed * time

substituting value of  speed and time in above equation we have

distance =  350/3 * 150 = 17,500 yards

Distance traveled by Austin's bike in 2.5 hours is 17,500 yards.

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Answer:

\boxed{\bold{120 \ Units^2}}

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Area Of Triangle: \bold{\frac{1}{2}Base \ \cdot \ Height}

Base: 10

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Insert Numbers Into Equation

\bold{\frac{1}{2}10 \ \cdot \ 24}

\bold{\frac{1}{2}} 10 = 5

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3 years ago
Rearrange the equation so U is the independent variable. 4u+8w=−3u+2w
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Answer:

u=-\frac{6}{7} w

Step-by-step explanation:

4u+8w=-3u+2w

Add 3u and -8w on both sides.

4u+3u=2w-8w

Add or subtract like terms.

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Divide 7 on both sides.

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3 years ago
What is the answer?<br> xy = 30<br> and x + y = 18
drek231 [11]

from second equation y = 18-x so:-

x(18 - x) = 30

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x = 16.14, 1.86


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3 0
2 years ago
Day care centers expose children to a wider variety of germs than the children would be exposed to if they stayed at home more o
Ostrovityanka [42]

Answer:

a) The study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is approximately 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is approximately 0.8565

c) The odds ratio is approximately 0.6780

d) The 95% CI is approximately 0.523 < OR < 0.833

e) i) Yes

ii) Children with more social activity have a higher occurrence of acute lymphoblastic leukemia

Step-by-step explanation:

a) An experimental study is a study in which the researcher adds inputs to the study and then monitors the outcomes

An observational study is one in which the researcher observes risk factors and does not intervene in the process

Therefore, the study is an observational study

b) The proportion of children with significant social activity in children with acute lymphoblastic leukemia is \hat p_1 = 1020/1272 = 85/106 = 0.8\overline {0188679245283} ≈ 0.802

The proportion of children with significant social activity in children without acute lymphoblastic leukemia is \hat p_2 = 5343/6238 ≈ 0.8565

c) We have the following two way table;

\begin{array}{ccc}Lymphoblastic \ Leukemia &With \ Social \ Activity & Without \ Social \ Activity\\With&1020 \ (a)&252 \ (b)\\Without&5343 \ (c)&895 \ (d) \end{array}

The \ odds \ ratio = \dfrac{1,020 \times 895}{5,343 \times 252} \approx 0.6780

The odds ratio ≈ 0.6780

d) The 95% confidence interval for the odds ratio is given as follows;

95 \% \ CI = OR \ \pm \ 1.96 \times \sqrt{\dfrac{1}{a} +\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}}

Where;

a = 1020, b = 252, c = 5343, d = 895

Therefore;

95 \% \ CI \approx  0.6780 \ \pm \ 1.96 \times \sqrt{\dfrac{1}{1,020} +\dfrac{1}{252}+\dfrac{1}{5,343}+\dfrac{1}{895}} \approx 0.678 \pm0.155

The 95% CI ≈ 0.523 < OR < 0.833

e) i) Given that the observed Odds Ratio is within the Confidence Interval, therefore, there is an indication that the amount of social activity is associated with acute lymphoblastic leukemia

ii) Based on the proportion of the study findings, children with more social activity have a higher occurrence of acute lymphoblastic leukemia

6 0
2 years ago
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