Which fraction names the shades part of the set

Which attteibutes do the parent cubic and cube root functions have ?

masya89 [10]

Answer:

its b duhhhhhhhhhh put me brianliest please pls lls

8 0

3 years ago

Give the expression 4n^2t/2n^t-1, write the expression as a binomial.

lubasha [3.4K]

binomial is an algebraic expression containing 2 terms. For example, (x + y) is a binomial.

We sometimes need to expand binomials as follows:

(a + b)0 = 1

(a + b)1 = a + b

(a + b)2 = a2 + 2ab + b2

(a + b)3 = a3 + 3a2b + 3ab2 + b3

(a + b)4 = a4 + 4a3b + 6a2b2 + 4ab3 + b4

(a + b)5 = a5 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + b5

Clearly, doing this by direct multiplication gets quite tedious and can be rather difficult for larger powers or more complicated expressions.

Pascal's Triangle

We note that the coefficients (the numbers in front of each term) follow a pattern. [This was noticed long before Pascal, by the Chinese.]

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

You can use this pattern to form the coefficients, rather than multiply everything out as we did above.

The Binomial Theorem

We use the binomial theorem to help us expand binomials to any given power without direct multiplication. As we have seen, multiplication can be time-consuming or even not possible in some cases.

Properties of the Binomial Expansion (a + b)nThere are \displaystyle{n}+{1}n+1 terms.The first term is an and the final term is bn.Progressing from the first term to the last, the exponent of a decreases by \displaystyle{1}1 from term to term while the exponent of b increases by \displaystyle{1}1. In addition, the sum of the exponents of a and b in each term is n.If the coefficient of each term is multiplied by the exponent of a in that term, and the product is divided by the number of that term, we obtain the coefficient of the next term.

7 0

2 years ago

A random sample of 10 parking meters in a resort community showed the following incomes for a day. Assume the incomes are normal

GenaCL600 [577]

Answer:

A 95% confidence interval for the true mean is [$3.39, $6.01].

Step-by-step explanation:

We are given that a random sample of 10 parking meters in a resort community showed the following incomes for a day;

Incomes (X): $3.60, $4.50, $2.80, $6.30, $2.60, $5.20, $6.75, $4.25, $8.00, $3.00.

Firstly, the pivotal quantity for finding the confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ ~ $t_n_-_1$

where, $\bar X$ = sample mean income = $\frac{\sum X}{n}$ = $4.70

s = sample standard deviation = $\sqrt{\frac{\sum (X-\bar X)^{2} }{n-1} }$ = $1.83

n = sample of parking meters = 10

$\mu$ = population mean

Here for constructing a 95% confidence interval we have used a One-sample t-test statistics because we don't know about population standard deviation.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-2.262 < $t_9$ < 2.262) = 0.95 {As the critical value of t at 9 degrees of

freedom are -2.262 & 2.262 with P = 2.5%}

P(-2.262 < $\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }$ < 2.262) = 0.95

P( $-2.262 \times {\frac{s}{\sqrt{n} } }$ < ${\bar X-\mu$ < $2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

P( $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ < $\mu$ < $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ) = 0.95

95% confidence interval for $\mu$ = [ $\bar X-2.262 \times {\frac{s}{\sqrt{n} } }$ , $\bar X+2.262 \times {\frac{s}{\sqrt{n} } }$ ]

= [ $4.70-2.262 \times {\frac{1.83}{\sqrt{10} } }$ , $4.70+ 2.262 \times {\frac{1.83}{\sqrt{10} } }$ ]

= [$3.39, $6.01]

Therefore, a 95% confidence interval for the true mean is [$3.39, $6.01].

The interpretation of the above result is that we are 95% confident that the true mean will lie between incomes of $3.39 and $6.01.

Also, the margin of error = $2.262 \times {\frac{s}{\sqrt{n} } }$

= $2.262 \times {\frac{1.83}{\sqrt{10} } }$ = 1.31

4 0

2 years ago

Expanded form for 48,243. (

Dima020 [189]

(4*10,000)+(8*1,000)+(2*100)+(4*10)+(3*1)

40,000+8,000+200+40+3=48,243

That is one way. You can also do this using exponential form!

Hope that helps

4 0

2 years ago

3. if kx² + 2x + k = -kx have equal roots, find the values of k.

crimeas [40]

Answer:

k = - $\frac{2}{3}$ , k = 2

Step-by-step explanation:

Using the discriminant Δ = b² - 4ac

The condition for equal roots is b² - 4ac = 0

Given

kx² + 2x + k = - kx ( add kx to both sides )

kx² + 2x + kx + k = 0 , that is

kx² + (2 + k)x + k = 0 ← in standard form

with a = k, b = 2 + k and c = k , thus

(2 + k)² - 4k² = 0 ← expand and simplify left side

4 + 4k + k² - 4k² = 0

- 3k² + 4k + 4 = 0 ( multiply through by - 1 )

3k² - 4k - 4 = 0 ← in standard form

(3k + 2)(k - 2) = 0 ← in factored form

Equate each factor to zero and solve for k

3k + 2 = 0 ⇒ 3k = - 2 ⇒ k = - $\frac{2}{3}$

k - 2 = 0 ⇒ k = 2

5 0

2 years ago