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Stels [109]
1 year ago
11

The sum of two integers is 67 the larger is 23 less than twice the smaller find the two integers

Mathematics
1 answer:
Alekssandra [29.7K]1 year ago
4 0

Answer:

<h2><em>Smaller Integer = 30</em></h2><h2><em>Larger integer = 37</em></h2>

Step-by-step explanation:

<h2><u>Conditions:</u></h2>
  • <em>There are </em><em>two integers</em><em> whose value are to bee found out.</em>
  • <em>Sum</em><em> of those two integers</em><em> is 63.</em>
  • <em>The </em><em>larger integer</em><em> is 23 </em><em>less</em><em> than</em><em> twice the smaller</em><em> one.</em>
<h2><u>To Find:</u></h2>
  • <em>The </em><em>two integers</em><em> in the given condition.</em>
<h2><u>Solution:</u></h2>

Let,

The smaller integer be <em>x</em>

<em>So,</em> larger integer will be = <em>2x - 23</em>

Given,

The sum of<em> </em>these two integers is = <em>67</em>

∴ According to the condition,

=> x + 2x - 23 = 67

  • <em>(Taking the like terms on each sides)</em>

=> x + 2x = 23 + 67

  • <em>(Adding the like terms)</em>

=> 3x = 90

  • <em>(Dividing each side with 3)</em>

=> x = 30

Then,

The smaller integer is = x = <em>30</em>

and, Larger integer is = 2x - 23 = <em>37</em>

<u>Hence,</u>

<u>The required smaller and larger integers are 30 and 37 respectively (Ans)</u>

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50 pts please help me due today links or incorrect answers will get reported
Bumek [7]

Using the normal distribution, the area underneath the shaded region <u>between the two z-scores</u> is given by:

C. 0.6766.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

Hence, for this problem, the area is the <u>p-value of z = 0.75 subtracted by the p-value of z = -1.3</u>.

Looking at the z-table, the p-values are given as follows:

  • z = 0.75: 0.7734.
  • z = -1.3: 0.0968.

Then:

0.7734 - 0.0968 = 0.6766.

Which means that option C is correct.

More can be learned about the normal distribution at brainly.com/question/15181104

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3 0
2 years ago
A punch glass is in the shape of a hemisphere with a radius of 5 cm. If the punch is being poured into the glass so that the cha
Galina-37 [17]

Answer:

28.27 cm/s

Step-by-step explanation:

Though Process:

  • The punch glass (call it bowl to have a shape in mind) is in the shape of a hemisphere
  • the radius r=5cm
  • Punch is being poured into the bowl
  • The height at which the punch is increasing in the bowl is \frac{dh}{dt} = 1.5
  • the exposed area is a circle, (since the bowl is a hemisphere)
  • the radius of this circle can be written as 'a'
  • what is being asked is the rate of change of the exposed area when the height h = 2 cm
  • the rate of change of exposed area can be written as \frac{dA}{dt}.
  • since the exposed area is changing with respect to the height of punch. We can use the chain rule: \frac{dA}{dt} = \frac{dA}{dh} . \frac{dh}{dt}
  • and since A = \pi a^2 the chain rule above can simplified to \frac{da}{dt} = \frac{da}{dh} . \frac{dh}{dt} -- we can call this Eq(1)

Solution:

the area of the exposed circle is

A =\pi a^2

the rate of change of this area can be, (using chain rule)

\frac{dA}{dt} = 2 \pi a \frac{da}{dt} we can call this Eq(2)

what we are really concerned about is how a changes as the punch is being poured into the bowl i.e \frac{da}{dh}

So we need another formula: Using the property of hemispheres and pythagoras theorem, we can use:

r = \frac{a^2 + h^2}{2h}

and rearrage the formula so that a is the subject:

a^2 = 2rh - h^2

now we can derivate a with respect to h to get \frac{da}{dh}

2a \frac{da}{dh} = 2r - 2h

simplify

\frac{da}{dh} = \frac{r-h}{a}

we can put this in Eq(1) in place of \frac{da}{dh}

\frac{da}{dt} = \frac{r-h}{a} . \frac{dh}{dt}

and since we know \frac{dh}{dt} = 1.5

\frac{da}{dt} = \frac{(r-h)(1.5)}{a}

and now we use substitute this \frac{da}{dt}. in Eq(2)

\frac{dA}{dt} = 2 \pi a \frac{(r-h)(1.5)}{a}

simplify,

\frac{dA}{dt} = 3 \pi (r-h)

This is the rate of change of area, this is being asked in the quesiton!

Finally, we can put our known values:

r = 5cm

h = 2cm from the question

\frac{dA}{dt} = 3 \pi (5-2)

\frac{dA}{dt} = 9 \pi cm/s// or//\frac{dA}{dt} = 28.27 cm/s

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Answer:

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Step-by-step explanation:

2.65 + 1.85 + 2.46 = 6.96/4 = 1.74lbs.

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