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lukranit [14]
2 years ago
6

A sports club needs to raise at least $500 by selling chocolate bars for $2.50 each. Sebastian wants to know how many chocolate

bars, c, the sports club must sell in order to reach its goal. Which of the following must be true about the inequality and the resulting graph?
Mathematics
2 answers:
Mice21 [21]2 years ago
6 0

they must sell at least  200 choc bars

c=choc bars

2.5c>or=500

divide both sides by 2.5

c>or=200

Reil [10]2 years ago
4 0

Answer:

80 to right

Step-by-step explanation:

I did the Quiz on Edge that is the answer

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In 1944 Elion joined the Burroughs Wellcome Laboratories (now part of GlaxoSmithKline (a company that makes prescription medicines)). There she was first the assistant and then the colleague of Hitchings, with whom she worked for the next four decades. Elion and Hitchings developed an array (variety) of new drugs that were effective against leukemia, autoimmune disorders, urinary-tract infections, gout, malaria, and viral herpes. Their success was due primarily to their innovative (characterized by new or unique) research methods. Rather than using the trial-and-error approach used by previous pharmacologists, Elion and Hitchings examined the difference between the biochemistry of normal human cells and that of cancer cells, bacteria, viruses, and other pathogens (disease-causing agents). They used this information to create drugs that could target a particular pathogen without harming the human host's normal cells. Their methods enabled them to eliminate much of the guesswork and wasted effort typical in previous drug research.
3 0
2 years ago
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A company manufacturing oil seals wants to establish X and R control charts on the process. There are 25 preliminary samples of
alisha [4.7K]

Answer:

A ) i) X control chart : upper limit = 50.475, lower limit = 49.825

    ii) R control chart : upper limit =  1.191, lower limit = 0

Step-by-step explanation:

A) Finding the control limits

grand sample mean = 1253.75 / 25 = 50.15

mean range = 14.08 / 25 = 0.5632

Based on  X control CHART

The upper control limit ( UCL ) =

grand sample mean + A2* mean range ) = 50.15 + 0.577(0.5632) = 50.475

The lower control limit (LCL)=

grand sample mean - A2 *  mean range = 50.15 - 0.577(0.5632) = 49.825

Based on  R control charts

The upper limit = D4 * mean range = 2.114 * 0.5632 = 1.191

The lower control limit = D3 * mean range = 0 * 0.5632 = 0  

B) estimate the process mean and standard deviation

estimated process mean = 50.15 = grand sample mean

standard deviation = mean range / d2  = 0.5632 / 2.326 = 0.2421

note d2 is obtained from control table

6 0
2 years ago
An observational study found that the amount of sleep an employee gets each night is associated with job performance. The correl
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Answer:

The correct option is D.) Causation cannot be determined from an observational study.

Step-by-step explanation:

The conclusion is not correct because

D.) Causation cannot be determined from an observational study.

Causation determined from an observational study is speculative and cannot be confirmed without data from a real experiment.

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2 years ago
Write H(t) for the amount spent in the United States on health care in year t, where t is measured in years since 2000. The rate
Kisachek [45]

Answer:

Check the explanation

Step-by-step explanation:

Write H(t) for the total sum spent in the United States on health care in year t, where t is measured in years since 2000.

The rate of increase of H(t) was projected to rise from $100 billion per year in 2000 to approximately $190 billion per year in 2010

(a) Find a linear model for the rate of change H'(t)

Slope of line is (190-100)/(2010-2000) = 90/10 = 9 billion dollars per yr

H'(t) - 190 = 9 (t-10) =

H'(t) = 9t + 100   billions dollars per year

(b) Given that $1,300 billion was spent on health care in the United States in 2000, find the function H(t).

H'(t) = 9t + 100   billions dollars per year   TAKING INTEGRATION..

H(t) = 9t^2/2 + 100t + C given H(0) = 1300

H(0) = 1300 = C

H(t) = 9t^2/2 + 100t + 1300 billion dollars.

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2 years ago
A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each on
Zielflug [23.3K]

Answer:

The 90% confidence interval using Student's t-distribution is (9.22, 11.61).

Step-by-step explanation:

Since we know the sample is not big enough to use a z-distribution, we use student's t-distribution instead.

The formula to calculate the confidence interval is given by:

\bar{x}\pm t_{n-1} \times s/\sqrt{n}

Where:

\bar{x} is the sample's mean,

t_{n-1} is t-score with n-1 degrees of freedom,

s is the standard error,

n is the sample's size.

This part of the equation is called margin of error:

s/\sqrt{n}

We know that:

n=28

\bar{x}=10.41

degrees of freedom = 27

1-\alpha = 0.90 \Rightarrow \alpha = 0.10

t_{n-1} = 1.703

s = 3.71

Replacing in the formula with the corresponding values we obtain the confidence interval:

\bar{x}\pm t_{n-1} \times s/\sqrt{n} = 10.41 \pm 1.70 \times 3.71/\sqrt{28} = (9.22, 11.61)

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2 years ago
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