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LuckyWell [14K]
3 years ago
7

Simon bought 3.5 pounds of Beef on sale for $1.25 per pound at HEB. If Simon gave the cashier $10.00, how much change would he r

eceive from this purchase?
Mathematics
1 answer:
Elenna [48]3 years ago
5 0

Answer:

5.625

Step-by-step explanation:

You might be interested in
The lengths of nails produced in a factory are normally distributed with a mean of 4.84 centimeters and a standard deviation of
Helga [31]

Answer:

Top 3%: 4.934 cm

Bottom 3%: 4.746 cm

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 = 4.84, \sigma = 0.05

Top 3%

Value of Z when Z has a pvalue of 1 - 0.03 = 0.97. So X when Z = 1.88.

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

1.88 = \frac{X - 4.84}{0.05}

X - 4.84 = 0.05*1.88

X = 4.934

Bottom 3%

Value of Z when Z has a pvalue of 0.03. So X when Z = -1.88.

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

-1.88 = \frac{X - 4.84}{0.05}

X - 4.84 = 0.05*(-1.88)

X = 4.746

8 0
3 years ago
a woman 1.65. tall stood 50m away from the foot of a tower,and observe that the angle of elevation of the top of the tower to be
ElenaW [278]

\sf\huge\underline{\star Solution:-}

\rightarrowLet the women's height be AE and distance between the women and tower be AC.

Also let the height of tower be BC.

\rightarrowNow, clearly it is forming a triangle.

So, in triangle ABC,

\rightarrow\sf{Tan50° \:= \: \frac{BC}{AC}}

\rightarrow\sf{1.19 \:= \: \frac{BC}{50}}

\rightarrow\sf{1.19 ×50\:= \: BC}

\rightarrow\sf{59.5m\:= \: BC}

\rightarrowHence, BC = 59.5m

So, BD(total height of the tower)\sf{ = \:BC+CD}

\sf{= \:59.5+1.65}

\sf{=\: 61.15m}

Therefore, total height of the tower =\sf\purple{ 61.15m.}

________________________________

Hope it helps you:)

5 0
2 years ago
Read 2 more answers
Evaluate the function at the given values of the independent variable. Simplify the results.
wlad13 [49]

answers \\ a. \:  \: 0 \\ b. \:  \:  \frac{ - 63}{8}  \\ c. \:  \:  {c}^{3}  -  {5c}^{2}  \\ d. \:  \:  {t}^{3}  + 25t +  {10t}^{2}  \\ please \: see \: the \: attached \: picture \\ for \: full \: solution \\ hope \: it \: helps \\ good \: luck \: on \: your \: assignment

8 0
2 years ago
Use the given minimum and maximum data​ entries, and the number of​ classes, to find the class​ width, the lower class​ limits,
Gennadij [26K]

Using proportions and the information given, it is found that:

  • The class width is of 14.375.
  • The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
  • The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

-------------------------

  • Minimum value is 19.
  • Maximum value is of 134.
  • There are 8 classes.
  • The classes are all of equal width, thus the width is of:

W = \frac{134 - 19}{8} = 14.375

-------------------------

The intervals will be of:

19 - 33.375

33.375 - 47.750

47.750 - 62.125

62.125 - 76.500

76.500 - 90.875

90.875 - 105.250

105.250 - 119.625

119.625 - 134.

  • The lower class limits are: {19, 33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625}.
  • The upper class limits are: {33.375, 47.750, 62.125, 76.500, 90.875, 105.250, 119.625, 134}.

A similar problem is given at brainly.com/question/16631975

6 0
3 years ago
What is 10÷3(4÷6)+.50
lianna [129]
52 2/9 is the answer.
8 0
3 years ago
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