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3 years ago

What is 21,321 rounded to the nearest 100

Mathematics

2 answers:

Vikentia [17]3 years ago

6 0

Answer:

Step-by-step explanation:

21,300

likoan [24]3 years ago

3 0

Answer:

21,300

That's rounding with the nearest hundredth

because 321 is neasrest 300 than 400

thats why the answer is

21,300

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Which fraction is closest to 1/2? A number line from negative 1 to 1 in halves. 1/6 3/8 3/4 ‐1/2

Arturiano [62]

Answer: 3/8

because 3/8 is .375
And 3/4 is .75
And 1/6 is .1666

1/2 is .50
3/8 is closest

5 0

3 years ago

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with mean of 1262 and a s

Andrew [12]

Answer:

a) 1186

b) Between 1031 and 1493.

c) 160

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Normally distributed with mean of 1262 and a standard deviation of 118.

This means that $\mu = 1262, \sigma = 118$

a) Determine the 26th percentile for the number of chocolate chips in a bag. 

This is X when Z has a p-value of 0.26, so X when Z = -0.643.

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

$-0.643 = \frac{X - 1262}{118}$

$X - 1262 = -0.643*118$

$X = 1186$

(b) Determine the number of chocolate chips in a bag that make up the middle 95% of bags.

Between the 50 - (95/2) = 2.5th percentile and the 50 + (95/2) = 97.5th percentile.

2.5th percentile:

X when Z has a p-value of 0.025, so X when Z = -1.96.

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

$-1.96 = \frac{X - 1262}{118}$

$X - 1262 = -1.96*118$

$X = 1031$

97.5th percentile:

X when Z has a p-value of 0.975, so X when Z = 1.96.

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

$1.96 = \frac{X - 1262}{118}$

$X - 1262 = 1.96*118$

$X = 1493$

Between 1031 and 1493.

(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip cookies?

Difference between the 75th percentile and the 25th percentile.

25th percentile:

X when Z has a p-value of 0.25, so X when Z = -0.675.

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

$-0.675 = \frac{X - 1262}{118}$

$X - 1262 = -0.675*118$

$X = 1182$

75th percentile:

X when Z has a p-value of 0.75, so X when Z = 0.675.

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

$0.675 = \frac{X - 1262}{118}$

$X - 1262 = 0.675*118$

$X = 1342$

IQR:

1342 - 1182 = 160

7 0

3 years ago

A circle is growing so that the radius is increasing at the rate of 3 cm/min. How fast is the area of the circle changing at the

Naya [18.7K]

Answer:

The area is growing at a rate of $\frac{dA}{dt} =226.2 \,\frac{cm^2}{min}$

Step-by-step explanation:

<em>Notice that this problem requires the use of implicit differentiation in related rates (some some calculus concepts to be understood), and not all middle school students cover such.</em>

We identify that the info given on the increasing rate of the circle's radius is 3 $\frac{cm}{min}$ and we identify such as the following differential rate:

$\frac{dr}{dt} = 3\,\frac{cm}{min}$

Our unknown is the rate at which the area (A) of the circle is growing under these circumstances,that is, we need to find $\frac{dA}{dt}$ .

So we look into a formula for the area (A) of a circle in terms of its radius (r), so as to have a way of connecting both quantities (A and r):

$A=\pi\,r^2$

We now apply the derivative operator with respect to time ( $\frac{d}{dt}$ ) to this equation, and use chain rule as we find the quadratic form of the radius:

$\frac{d}{dt} [A=\pi\,r^2]\\\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}$

Now we replace the known values of the rate at which the radius is growing ( $\frac{dr}{dt} = 3\,\frac{cm}{min}$ ), and also the value of the radius (r = 12 cm) at which we need to find he specific rate of change for the area :

$\frac{dA}{dt} =\pi\,*2*r*\frac{dr}{dt}\\\frac{dA}{dt} =\pi\,*2*(12\,cm)*(3\,\frac{cm}{min}) \\\frac{dA}{dt} =226.19467 \,\frac{cm^2}{min}\\$