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kobusy [5.1K]
3 years ago
10

You and a friend both would like a salad and a small drink. Between the two of you, you have $8.00. A salad costs $2.49 and a sm

all drink is $0.99. Can either of you have a second salad or a drink? Assume there is No Sales Tax!
A- Yes, one Salad
B- Yes, one of Each
C- Yes, one Drink
D- No, you can't
Mathematics
2 answers:
kolezko [41]3 years ago
8 0
C - Yes, one Drink. Is the answer
Gennadij [26K]3 years ago
5 0

Answer: System of Linear Inequalities:

1. C. yes, one drink

2. D. -2x+y>4 x+y<2

3. C. 3

4. D 2/3

posting the answers along with the letters just in case if it's not the same for you, you are still able to get a good grade hopefully :)

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Name a pair of perpendicular lines: <br>is line BC and AJ perpendicular: explain ​
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A pair of perpendicular lines would be line CF and line AJ.

Line BC and line AJ are NOT perpendicular because when a pair of lines are perpendicular, all angles become 90 degrees. It is obvious, in the case, if you were to slide line BC on top of line AJ, the angles will not equal 90 degrees.
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A point has a positive x and y coordinate- answer is Quadrant I
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A chocolate bar contains 6 grams of fat. If you eat no fewer than 5 and no more than 9 bars, how many grams of fat will you cons
IRINA_888 [86]

5 x 6 = 30

9 x 6 = 54

54 + 30 = 84

you would be intaking 84 grams of fat

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2 years ago
The average size of 8 farms in Indiana County, Pennsylvania, is 191 acres with standard deviation of 38 acres. The average size
Grace [21]

Answer:

Step-by-step explanation:

This is a test of 2 independent groups. The population standard deviations are not known. it is a two-tailed test. Let μ1 be average size of farms in Indiana County, Pennsylvania and μ2 be the average size of farms in Greene County, Pennsylvania.

The random variable is μ1 - μ2 = difference in the average size of the farms in Indiana County and the average size of the farms in Greene County.

We would set up the hypothesis.

The null hypothesis is

H0 : μ1 = μ2 H0 : μ1 - μ2 = 0

The alternative hypothesis is

H1 : μ1 ≠ μ2 H1 : μ1 - μ2 ≠ 0

Since sample standard deviation is known, we would determine the test statistic by using the t test. The formula is

(μ1 - μ2)/√(s1²/n1 + s2²/n2)

From the information given,

μ1 = 191

μ2 = 199

s1 = 38

s2 = 12

n1 = 8

n2 = 10

t = (191 - 199)/√(38²/8 + 12²/10)

t = - 0.57

The formula for determining the degree of freedom is

df = [s1²/n1 + s2²/n2]²/(1/n1 - 1)(s1²/n1)² + (1/n2 - 1)(s2²/n2)²

df = [38²/8 + 12²/10]²/[(1/8 - 1)(38²/8)² + (1/10 - 1)(12²/10)²] = 37986.01/4677.36142857143

df = 8

We would determine the probability value from the t test calculator. It becomes

p value = 0.58

Let us assume a significance level of 0.05

Since alpha, 0.05 < than the p value, 0.58, then we would accept the null hypothesis. Therefore, at a significance level of 5%, there is not sufficient evidence to conclude that the average size of the farms in Indiana County and the average size of the farms in Greene County are different.

4 0
3 years ago
A researcher wanted to see if giving a selective serotonin reuptake inhibitor (anti-depressant) would decrease the number of sel
Delvig [45]

Answer:

a. The mean of the sample is M=35.

The variance of the sample is s^2=39.125.

The standard deviation of the sample is s=6.255.

b. z=-1.6

c. SEM = 2.212

Step-by-step explanation:

The mean of the sample is M=35.

The variance of the sample is s^2=39.125.

The standard deviation of the sample is s=6.255.

<u>Sample mean</u>

<u />M=\dfrac{1}{8}\sum_{i=1}^{8}(27+25+32+40+43+37+35+38)\\\\\\ M=\dfrac{277}{8}=35

<u>Sample variance and standard deviation</u>

<u />s^2=\dfrac{1}{(n-1)}\sum_{i=1}^{8}(x_i-M)^2\\\\\\s^2=\dfrac{1}{7}\cdot [(27-(35))^2+(25-(35))^2+(32-(35))^2+(40-(35))^2+(43-(35))^2+(37-(35))^2+(35-(35))^2+(38-(35))^2]\\\\\\

s^2=\dfrac{1}{7}\cdot [(58.141)+(92.641)+(6.891)+(28.891)+(70.141)+(5.64)+(0.14)+(11.39)]\\\\\            s^2=\dfrac{273.875}{7}=39.125\\\\\\s=\sqrt{39.125}=6.255

b. If the population mean is 45, the z-score for M=35 would be:

z=\dfrac{M-\mu}{\sigma}=\dfrac{35-45}{6.255}=\dfrac{-10}{6.255}=-1.6

c. The standard error of the mean (SEM) of this group is calculated as:

SEM=\dfrac{s}{\sqrt{n}}=\dfrac{6.255}{\sqrt{8}}=\dfrac{6.255}{2.828}=2.212

4 0
3 years ago
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