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oksian1 [2.3K]
3 years ago
7

Patricia had less than $9 to spend at the grocery store. she bought 1 pound of tomatoes and wanted to spend the rest of her mone

y on potatoes. one pound of tomatoes cost $2.40 and one pound of potatoes cost $2.20
Mathematics
2 answers:
iragen [17]3 years ago
6 0

Answer:

$2.40 + $2.20x < $9

Step-by-step explanation:

Patricia had less than $9 to spend at the grocery store.

She bought 1 pound of tomatoes and wanted to spend the rest amount on potatoes.

Since 1 pound of tomatoes cost = $2.40

If 1 pound of potatoes cost = $2.20

Now cost of x pounds of potatoes = $(2.20).(x)

Total money spent = money spent on potatoes + amount spent on potatoes

$9 = $2.40 + $(2.20).(x)

But Patricia is having money less than $9 to spend, so total amount spent

will be less than $9 and the inequality will be as below.

$9 > $2.40 + $(2.20)(x)

shepuryov [24]3 years ago
3 0
2.40+2.20=4.60

9-4.60=4.40

Patricia has $4.40 left to spend (or save)
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For a medical study, a researcher wishes to select people in the middle 60% of the population based on blood pressure.
Finger [1]

Answer:

Lower limit: 113.28

Upper limit: 126.72

Step-by-step explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

\mu = 120, \sigma = 8

Middle 60%

So it goes from X when Z has a pvalue of 0.5 - 0.6/2 = 0.2 to X when Z has a pvalue of 0.5 + 0.6/2 = 0.8

Lower limit

X when Z has a pvalue of 0.20. So X when Z = -0.84

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

-0.84 = \frac{X - 120}{8}

X - 120 = -0.84*8

X = 113.28

Upper limit

X when Z has a pvalue of 0.80. So X when Z = 0.84

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

0.84 = \frac{X - 120}{8}

X - 120 = 0.84*8

X = 126.72

4 0
3 years ago
What is one of the solutions to the following equations? y2+x2=53<br> Y-x=5
Olenka [21]
(1)  y² + x² = 53
(2)  y - x = 5 ⇒ y = x + 5

subtitute (2) to (1)

(x + 5)² + x² = 53      |use (a + b)² = a² + 2ab + b²

x² + 2x·5 + 5² + x² = 53

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x² + 5x - 14 = 0

x² - 2x+ 7x - 14 = 0

x(x - 2) + 7(x - 2) = 0

(x - 2)(x + 7) = 0 ⇔ x - 2 = 0 or x + 7 = 0 ⇔ x = 2 or x = -7

subtitute the values of y to (2)

for x = 2, y = 5 + 2 = 7
for x = -7, y = 5 + (-7) = 5 - 2 = 3

Answer: x = 2 and y = 7 or x = -7 and y = 3
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Answer:

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