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2 years ago

Please help i’m literally so stressed out

1 answer:

3 0

The inequality that can be used to determine x, the maximum number of outfits Luis can purchase while staying within his budget is x ≤ 4

<h3>How to solve inequality?</h3>

Luis has $420 to spend at a bicycle store for some new gear and biking outfits.

He buys a new bicycle for $209.67.

He buys 2 bicycle reflectors for $6.04 each and a pair of bike gloves for $25.77.

He plans to spend some or all of the money he has left to buy new biking outfits for $37.16 each.

Therefore,

x = maximum number of outfits Luis can purchase while staying within his budget.

Hence, the inequality that can be used to represent the situation is as follows;

209.67 + 6.04(2) + 25.77 + 37.16x ≤ 420

235.44 + 12.08 + 37.16x ≤ 420

247.52 + 37.16x ≤ 420

37.16x ≤ 420 - 247.52

37.16x ≤ 172.48

divide both sides by 37.16

x ≤ 4.64155005382

x ≤ 4

Therefore, the inequality that can be used to determine x, the maximum number of outfits Luis can purchase while staying within his budget is x ≤ 4

learn more on inequality here: brainly.com/question/24565096

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No solution.

Step-by-step explanation:

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3 years ago

What is the product of 3 x4/5

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Answer:

2 2/5

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8 0

3 years ago

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3 1/4 - 1 3/5 | please help !!

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Answer:

Step-by-step explanation:

Step: 1: Convert mixed fraction to improepr fraction.

Step: 2 : Find the LCM of 4 , 5 = 20

Step 3: Find equivalent fractions with denominator as 20

Step 4: Now subtract.

$3\dfrac{1}{4}-1\dfrac{3}{5}=\dfrac{13}{4}-\dfrac{8}{5}$

$\sf = \dfrac{13*5}{4*5}-\dfrac{8*4}{5*4}\\\\\\ = \dfrac{65}{20}-\dfrac{32}{20}\\\\\\ = \dfrac{65-32}{20}\\\\\\ = \dfrac{33}{20}\\\\ \\ = 1 \dfrac{13}{20}$

6 0

2 years ago

The caffeine content (in mg) was examined for a random sample of 50 cups of black coffee dispensed by a new machine. The mean an

Genrish500 [490]

Answer:

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 mg .

Step-by-step explanation:

Given -

The sample size is large then we can use central limit theorem

n = 50 ,

Standard deviation $(\sigma)$ = 7.1

Mean $\overline{(y)}$ = 110

$\alpha =$ 1 - confidence interval = 1 - .98 = .02

$z_{\frac{\alpha}{2}}$ = 2.33

98% confidence interval for the mean caffeine content for cups dispensed by the machine = $\overline{(y)}\pm z_{\frac{\alpha}{2}}\frac{\sigma}\sqrt{n}$

= $110\pm z_{.01}\frac{7.1}\sqrt{50}$

= $110\pm 2.33\frac{7.1}\sqrt{50}$

First we take + sign

$110 + 2.33\frac{7.1}\sqrt{50}$ = 112.34

now we take - sign

$110 - 2.33\frac{7.1}\sqrt{50}$ = 107.66

We are 98% confident interval for the mean caffeine content for cups dispensed by the machine between 107.66 and 112.34 .

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4 years ago