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natima [27]
3 years ago
10

Jaya forgets how to write numbers in expanded form. Explain the steps to her using the number 13.652 as an example, and then wri

te 13.652 in expanded form.
Mathematics
2 answers:
Amanda [17]3 years ago
7 0
In the expanded form the digits of the number are split into each of the individual digits with their place value and written in expanded form. The example of standard form of a number is 4,982 and the same number can be written in expanded form as 4 × 1000 + 9 × 100 + 8 × 10 + 2 × 1 = 4000 + 900 + 80 + 2.
miss Akunina [59]3 years ago
3 0

Answer:

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Step-by-step explanation:

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If B = w + 10 and C = -w - 1, find an expression that equals B + 2C in standard form
enot [183]

Answer:

Step-by-step explanation:

B + 2C

= w + 10 + 2(-w - 1)

= w + 10 -2w - 2

= -w + 8

8 0
2 years ago
Find all points having an x-coordinate of 2 whose distance from the point (-1,-2) is 5
mariarad [96]
The\ equation\ of\ the\ circle:(x-a)^2+(y-b)^2=r^2\\\\where\ (a;\ b)\ -the\ coordinates\ of\ the\ center;\ r-the\ radius\\-----------------------------\\(-1;-2)-the\ center\ of\ the\ circle\\r=5\\\\The\ equation:(x+1)^2+(y+2)^2=5^5\\\\Put\ x=2\ to\ the\ equation:\\\\(2+1)^2+(y+2)^2=25\\3^2+(y+2)^2=25\\9+(y+2)^2=25\\(y+2)^2=25-9\\(y+2)^2=16\iff y+1=\pm\sqrt{16}\\\\y+1=-4\ or\ y+1=4\ \ \ \ \ |subtract\ 1\ from\ both \sides\\y=-5\ or\ y=3\\\\Answer:(2;-5)\ and\ (2;\ 3).
5 0
3 years ago
The temple at the top of pyramid is approximately 24 meters above the ground, and there are 91 steps leading up the temple. How
slamgirl [31]

Answer:

13.187 meters

Step-by-step explanation:

Basically, there are 24 meters for every 91 steps: \frac{24}{91}

There are x meters for every 50 steps: \frac{x}{50}

\frac{24}{91} =\frac{x}{50}

Cross multiply

91x = 1200

Divide both sides by 91 to isolate the variable, x

91x/91 = 1200/91

x = 13.186813

7 0
3 years ago
Unknown to a medical researcher, 6 out of 25 patients have a heart problem that will result in death if they receive the test dr
Fiesta28 [93]

Using the hypergeometric distribution, it is found that there is a 0.0002 = 0.02% probability that exactly 6 patients will die.

<h3>What is the hypergeometric distribution formula?</h3>

The formula is:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • N is the size of the population.
  • n is the size of the sample.
  • k is the total number of desired outcomes.

The values of the parameters for this problem are:

N = 25, k = 6, n = 8.

The probability that exactly 6 patients will die is P(X = 6), hence:

P(X = x) = h(x,N,n,k) = \frac{C_{k,x}C_{N-k,n-x}}{C_{N,n}}

P(X = 6) = h(6,25,8,6) = \frac{C_{6,6}C_{19,2}}{C_{25,8}} = 0.0002

0.0002 = 0.02% probability that exactly 6 patients will die.

More can be learned about the hypergeometric distribution at brainly.com/question/24826394

#SPJ1

8 0
2 years ago
The total mass of the Sun is about 2×10^30 kg, of which about 76 % was hydrogen when the Sun formed. However, only about 12 % of
nignag [31]

Answer:

A. 1.8 ×10^{30} Kg

B i. 3.0 × 10^{17} seconds

  ii. 9.6 × 10^{9} years

C. After 9.2 × 10^{9} (9.2 billion) years

Step-by-step explanation:

Given that the mass of the Sun = 2× 10^{30} Kg.

Mass of hydrogen when Sun was formed = 76% of 2× 10^{30} Kg

                            = \frac{76}{100}  ×2× 10^{30} Kg

                           = 1.52 × 10^{30} Kg

Mass of hydrogen available for fusion = 12% of 1.52 × 10^{30} Kg

                           = \frac{12}{100} × 1.52 × 10^{30} Kg

                           = 1.824 ×10^{30} Kg

A. Total mass of hydrogen available for fusion over the lifetime of the sun is 1.8 ×10^{30} Kg.

B. Given that the Sun fuses 6 × 10^{11} Kg of hydrogen each second.

i. The Sun's initial hydrogen would last;

                                     \frac{1.8*10^{30} }{6*10^{11} }

                                 = 3.04 × 10^{17} seconds

The Sun's hydrogen would last 3.0 × 10^{17} seconds

ii. Since there are 31536000 seconds in a year, then;

The Sun's initial hydrogen would last;

                                     \frac{3.04*10^{17} }{31536000}

                                 = 9.640 × 10^{9} years

The Sun's hydrogen would last 9.6 × 10^{9} years.

C. Given that our solar system is now about 4.6 × 10^{9} years, then;

                               \frac{9.6*10^{9} }{4.6*10^{9} }

                             = 2.09

So that;   2 × 4.6 × 10^{9} = 9.2 × 10^{9} years

Therefore, we need to worry about the Sun running out of hydrogen for fusion after 9.2 × 10^{9} years.

6 0
3 years ago
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