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3 years ago

The following table represents the sample space of rolling two number cubes.

Mathematics

2 answers:

devlian [24]3 years ago

6 0

Answer:

i think its ther 5 one

Step-by-step explanation:

i think

alex41 [277]3 years ago

4 0

We just need to count the number of outcomes that have exactly one even number. The total number of outcomes is 36. We see that from the first row, the number of outcomes that qualify are 3: 1-2, 1-4, 1-6 (out of the 6 total outcomes). For the 2nd row, again there are 3: 2-1, 2-3, 2-5. It is obvious by now that for any row, whether the number is even or odd, there are 3 outcomes on that row that correspond to rolling exactly one even number. Hence, there are in total 6*3=18 favorable outcomes.

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What fraction of 5 is 1/2

Sphinxa [80]

Answer:

1/10

Step-by-step explanation:

1/2 divided by 5 = 1/10

7 0

3 years ago

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HELP NEEDED !!!!

Fofino [41]

The answer is Letter D - 84%

68% of the data is within +/- 1 SD from the mean = 59 to 7195% of the data is within +/- 2 SDs from the mean = 53 to 7799.7% of the data is within +/- 3 SDs from the mean = 47 to 83

The whole 68% (59 - 71) scored below 71. Since the data are evenly distributed, we'll get the 32% (100 - 68) outside the range of 68% and divide it by 2 and add the quotient to the original 68% so that we'll get the total data below 71.
100 - 68 = 32
32 / 2 = 16
16 + 68 = 84

7 0

3 years ago

Q.5(b) The population {(P) in millions} of a country is estimated by the function, P=125e0.035t, t = time measured in years sinc

nekit [7.7K]

Answer:

Hello,

Step-by-step explanation:

Q.5(b) The population {(P) in millions} of a country is estimated by the function, P=125e0.035t, t = time measured in years since 1990. (a) what is the population expected to equal in year 2000 (b) determine the expression for the instantaneous rate of change in the population (c) what is the instantaneous rate of change in the population expected to equal in year 2000.

$P(t)=125*e^{0.035*t}\\a)\\P(2000)=125*e^{0.035*(2000-1990)}=177.38...\\\\b)\\P'(t)=125*0.035*e^{0.035*t}\\\\c)\\\\P'(2000)=125*0.035*e^{0.035*(2000-1990)}=6.2084...$

5 0

3 years ago

Find number a and k so that x-2 is afactor<br>OF F(x)=x4-2ax tax-<br>X+k and F(-1)=3

Slav-nsk [51]

Question:

Find numbers a and k so that x-2 is a factor of

$f(x)=x^4-2ax^3+ax^2- x+k$

and

$f(-1)=3$

Answer:

$k = -2$ and $a=1$

Step-by-step explanation:

Given

$f(x)=x^4-2ax^3+ax^2- x+k$

$Factor:\ x - 4$

$f(-1)=3$

Required

Find a and k

For $f(-1)=3$

Substitute -1 for x

$f(x)=x^4-2ax^3+ax^2- x+k$

$f(-1) = (-1)^4 - 2a *(-1)^3 + a*(-1)^2 - (-1) + k$

$f(-1) = 1 - 2a *-1 + a*1 +1 + k$

$f(-1) = 1 +2a + a +1 + k$

$f(-1) = 2 +3a + k$

Substitute 3 for f(-1)

$3 = 2 +3a + k$

Collect Like Terms

$3 - 2 = 3a + k$

$1 = 3a + k$

Also:

If $x - 2$ is a factor, then

$f(2) = 0$

Substitute 2 for x and 0 for f(x)

$f(x)=x^4-2ax^3+ax^2- x+k$

$0 = 2^4 - 2a * 2^3 + a * 2^2 - 2 + k$

$0 = 16 - 2a * 8 + a * 4 - 2 + k$

$0 = 16 - 16a + 4a - 2 + k$

$0 = 16 - 12a - 2 + k$

Collect Like Terms

$2 - 16 = - 12a + k$

$-14 = k - 12a$

$k = 12a - 14$

Substitute 12a - 14 for k in $1 = 3a + k$

$1 = 3a + 12a - 14$

$1 = 15a - 14$

Collect Like Terms

$15a = 1 + 14$

$15a = 15$

Solve for a

$a = \frac{15}{15}$

$a=1$

Substitute 1 for a in $k = 12a - 14$

$k = 12 * 1 - 14$

$k = 12 - 14$

$k = -2$

3 0

2 years ago

What does product rule of exponents say

LuckyWell [14K]

The rules of exponents are
$( x^{m})( x^{n})=( x^{m+n})$
$\frac{ x^{m} }{ x^{n} } = x^{m-n}$
$x^{ \frac{m}{n} } = \sqrt[n]{ x^{m} }$
$( x^{m} )( y^{m} )=(yx)^{m}$
$( x^{m})^{n}=x^{mn}$
$( \frac{x}{y})^{m}= \frac{x^{m}}{y^{m}}$
$x^{-n}= \frac{1}{x^{n}}$
these are just some basic ones

7 0

3 years ago

Read 2 more answers