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3 years ago

There are men and women at a meeting.

Mathematics

2 answers:

TEA [102]3 years ago

7 0

Answer:

70 people

Step-by-step explanation:

<h2>solution</h2><h2>40% are men </h2><h2>take 100%_40%=60% for women</h2><h2>equate if 60% =42 women therefore 100% will be</h2><h2>100%×42 divide by 60%=70 people</h2>

Usimov [2.4K]3 years ago

5 0

Answer:

70 people are present

Step-by-step explanation:

i just started at 100 and went down solving each one for 40 % and trying to find the other 60 and 60 % of 70 was 42

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What is hypotenuse of triangle

aev [14]

The side opposite the right angle.

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2 years ago

Let p be the population proportion for the following condition. Find the point estimates for p and q. In a survey of 1226 adults

weeeeeb [17]

Answer:

(a)0.3303

(b)0.6697

Step-by-step explanation:

Point estimation is the use of sample data to calculate a single value which serves as a best estimate of a given unknown population parameter.

Total Sample=1226

Number who said that they were not confident that the food they eat in country A is safe=405.

The point estimate of those who said that they were not confident that the food they eat in country A is safe is given as $\^{p}$

(a)Point Estimate for p,

$\^{p}=\frac{405}{1226} =0.3303$

(b)Point estimate for q, $\^{q}=$ 1- (point estimate for p)

=1-0.3303

=0.6697

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2 years ago

change in the position of the hand with the angle made by the change. The minute hand moves from 2 to 9. The minute hand moves f

lys-0071 [83]

Answer:

2–9 7π/6
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4–6 π/3
1–4 π/2
5–10 5π/6

Step-by-step explanation:

The angle between numbers on an analog clock is π/6 radians. So, the angles of interest are ...

2 to 9 : seven numbers : 7π/6 radians

3 to 7 : four numbers : 4π/6 = 2π/3 radians

4 to 6 : two numbers : 2π/6 = π/3 radians

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5 to 10 : five numbers : 5π/6 radians

4 0

3 years ago

Dale has $25 to spend at the Fresno fair. If the admission is $8 and the rides cost $2 each, what is the greatest number of ride

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Step-by-step explanation:

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2 years ago

P(x) = x + 1x² – 34x + 343<br> d(x)= x + 9

Feliz [49]

Answer:

$x=\frac{9}{d-1},\:P=\frac{-297d+378}{\left(d-1\right)^2}+343$

Step-by-step explanation:

Let us start by isolating x for dx = x + 9.

dx - x = x + 9 - x > dx - x = 9.

Factor out the common term of x > x(d - 1) = 9.

Now divide both sides by d - 1 > $\frac{x\left(d-1\right)}{d-1}=\frac{9}{d-1};\quad \:d\ne \:1$ . Go ahead and simplify.

$x=\frac{9}{d-1};\quad \:d\ne \:1$ .

Now, $\mathrm{For\:}P=x+1x^2-34x+343, \mathrm{Subsititute\:}x=\frac{9}{d-1}$ .

$P=\frac{9}{d-1}+1\cdot \left(\frac{9}{d-1}\right)^2-34\cdot \frac{9}{d-1}+343$ .

Group the like terms... $1\cdot \left(\frac{9}{d-1}\right)^2+\frac{9}{d-1}-34\cdot \frac{9}{d-1}+343$ .

$\mathrm{Add\:similar\:elements:}\:\frac{9}{d-1}-34\cdot \frac{9}{d-1}=-33\cdot \frac{9}{d-1}$ > $1\cdot \left(\frac{9}{d-1}\right)^2-33\cdot \frac{9}{d-1}+343$ .

Now for $1\cdot \left(\frac{9}{d-1}\right)^2 > \mathrm{Apply\:exponent\:rule}: \left(\frac{a}{b}\right)^c=\frac{a^c}{b^c} > \frac{9^2}{\left(d-1\right)^2} = 1\cdot \frac{9^2}{\left(d-1\right)^2}$ .

$\mathrm{Multiply:}\:1\cdot \frac{9^2}{\left(d-1\right)^2}=\frac{9^2}{\left(d-1\right)^2}$ .

Now for $33\cdot \frac{9}{d-1} > \mathrm{Multiply\:fractions}: \:a\cdot \frac{b}{c}=\frac{a\:\cdot \:b}{c} > \frac{9\cdot \:33}{d-1} > \frac{297}{d-1}$ .

Thus we then get $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}+343$ .

Now we want to combine fractions. $\frac{9^2}{\left(d-1\right)^2}-\frac{297}{d-1}$ .

$\mathrm{Compute\:an\:expression\:comprised\:of\:factors\:that\:appear\:either\:in\:}\left(d-1\right)^2\mathrm{\:or\:}d-1 > This\: is \:the\:LCM > \left(d-1\right)^2$

$\mathrm{For}\:\frac{297}{d-1}:\:\mathrm{multiply\:the\:denominator\:and\:numerator\:by\:}\:d-1 > \frac{297}{d-1}=\frac{297\left(d-1\right)}{\left(d-1\right)\left(d-1\right)}=\frac{297\left(d-1\right)}{\left(d-1\right)^2}$

$\frac{9^2}{\left(d-1\right)^2}-\frac{297\left(d-1\right)}{\left(d-1\right)^2} > \mathrm{Since\:the\:denominators\:are\:equal,\:combine\:the\:fractions}> \frac{a}{c}\pm \frac{b}{c}=\frac{a\pm \:b}{c}$

$\frac{9^2-297\left(d-1\right)}{\left(d-1\right)^2} > 9^2=81 > \frac{81-297\left(d-1\right)}{\left(d-1\right)^2}$ .

Expand $81-297\left(d-1\right) > -297\left(d-1\right) > \mathrm{Apply\:the\:distributive\:law}: \:a\left(b-c\right)=ab-ac$ .

$-297d-\left(-297\right)\cdot \:1 > \mathrm{Apply\:minus-plus\:rules} > -\left(-a\right)=a > -297d+297\cdot \:1$ .

$\mathrm{Multiply\:the\:numbers:}\:297\cdot \:1=297 > -297d+297 > 81-297d+297 > \mathrm{Add\:the\:numbers:}\:81+297=378 > -297d+378 > \frac{-297d+378}{\left(d-1\right)^2}$

Therefore $P=\frac{-297d+378}{\left(d-1\right)^2}+343$ .

Hope this helps!

5 0

3 years ago