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katen-ka-za [31]
1 year ago
6

Estimate the value 9.9 squared times 1.79 and the square root of 97.5 divided by 1.96

Mathematics
1 answer:
Inessa05 [86]1 year ago
8 0

Answer:

200 and 5

Step-by-step explanation:

1. 9.9 rounded is 10

1.79 rounded is 2

10^2 is 100

100 x 2 = 200

2. 97.5 rounded is 100

1.96 rounded is 2

square root of 100 is 10

10/2 = 5

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(5.5x + 6.2y) + (4.3x + 8.3z) + (1.6z - 5.ly)
Combine Like Terms
9.8x + 1.1y + 14.5z 

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9.8x + 1.1y + 14.5z
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Roberta invested $600 into a mutual fund that paid 4% interest each year compounded annually. Write an exponential function of t
Oksanka [162]

Answer:

The exponential equation is <em>A = 600(1.04)^15</em>

<em></em>

The value of the mutual fund after 15 years is <em>$1,081</em>

Step-by-step explanation:

The  value of the mutual fund after the number of years can be represented using the compound interest  equation below;

A = P(1 + r/n)^nt

Where A is the value of the mutual fund after 15 years, P is the initial amount invested which is $600, r is the interest rate which is 4% or 0.04(4% = 4/100 = 0.04), n is the number of times we are compounding per year(which is 1 since it is a one time payment per year) and t is the number of years which is 15

Let's plug these values, we have;

A = 600(1 + 0.04/1)^15

A = 600(1.04)^15

A = $1,081 approximately

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Step-by-step explanation:

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Help me solve this . <br> -2x + 4 &lt; 24
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A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints
Sliva [168]

Answer:

a)0.099834

b) 0

Step-by-step explanation:

To solve for this question we would be using , z.score formula.

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.

a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.

Standard Deviation = √variance

= √0.16 = 0.4

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x = 20.857

z = (x-μ)/σ

z = 20.857 - 21.37/0.4

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P(x<20.857) = 0.099834

b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.

z score formula used = (x-μ)/σ/√n

x = 20.857

Standard deviation = 0.4

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n = 100

z = 20.857 - 21.37/0.4/√100

= 20.857 - 21.37/ 0.4/10

= 20.857 - 21.37/ 0.04

= -12.825

P-value from Z-Table:

P(x<20.857) = 0

c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.

5 0
1 year ago
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