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2 years ago

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Annette is taller than Heather but shorter than garo Tanya's height is between Garo and Annette's Karin would be shortest if it

weren't for Alexa list the names in order from shortest to tallest

1 answer:

nata0808 [166]2 years ago

6 0

I think it would be Garo, Annette, and Heather????

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Can someone help explain this?

bogdanovich [222]

The mode is whichever one has the most so the mode would be 7. And the range is the highest number to the lowest. So 9-4 which is 5

3 0

1 year ago

4, 16, 64, 256, 1024, ... what is the exponential function

muminat

Answer:

y=x^4

Step-by-step explanation:

the equation is multiplying by 4 for every x value

5 0

2 years ago

An e-commerce research company claims that 60% or more graduate students have bought merchandise on-line at their site. A consum

Rasek [7]

Answer:

We accept H₀ we don´t have enough evidence to conclude that a consumer group position is correct

Step-by-step explanation:

We have a case of test of proportion, as a consumer group is suspicious of the claim and think the proportion is lower we must develop a one tail test (left tail) Then

1.- Test hypothesis:

Null hypothesis H₀ P = P₀

Alternative hypothesis Hₐ P < P₀

2.- At significance level of α = 0,05 Critical value

z(c) = -1,64

3.-We compute z(s) value as:

z(s) = ( P - P₀ )/ √P*Q/n where P = 44/80 P = 0,55 and Q = 0,45

P₀ = 0,6 and n = 80

Plugging all these values in the equation we get:

z(s) = ( 0,55 - 0,6 ) / √(0,2475/80)

z(s) = - 0,05/ √0,0031

z(s) = - 0,05/0,056

z(s) = - 0,8928

4.-We compare z(s) and z(c)

z(s) > z(c) -0,8928 on the left side it means that z(s) is in the acceptance region so we accept H₀

7 0

2 years ago

Please help!!! Express the number in terms of i

Effectus [21]

Answer:

6i sqrt 2

Step-by-step explanation:

3 0

2 years ago

Help me with solution please =(

Slav-nsk [51]

$\huge \boxed{\mathfrak{Question} \downarrow}$

Factorise the polynomials.

$\large \boxed{\mathfrak{Answer \: with \: Explanation} \downarrow}$

<h3><u>1. b² + 8b + 7</u></h3>

$\sf \: b ^ { 2 } + 8 b + 7$

Factor the expression by grouping. First, the expression needs to be rewritten as b²+pb+qb+7. To find p and q, set up a system to be solved.

$\sf \: p+q=8 \\ \sf pq=1\times 7=7$

As pq is positive, p and q have the same sign. As p+q is positive, p and q are both positive. The only such pair is the system solution.

$\sf \: p=1 \\ \sf \: q=7$

Rewrite $\sf\:b^{2}+8b+7 \: as \: \left(b^{2}+b\right)+\left(7b+7\right)$ .

$\sf \: \left(b^{2}+b\right)+\left(7b+7\right)$

Take out the common factors.

$\sf \: b\left(b+1\right)+7\left(b+1\right)$

Factor out common term b+1 by using distributive property.

$\boxed{\boxed{ \bf \: \left(b+1\right)\left(b+7\right) }}$

__________________

<h3><u>2. 4x² + 4x + 1</u></h3>

$\sf \: 4 x ^ { 2 } + 4 x + 1$

Factor the expression by grouping. First, the expression needs to be rewritten as 4x²+ax+bx+1. To find a and b, set up a system to be solved.

$\sf \: a+b=4 \\ \sf ab=4\times 1=4$

As ab is positive, a and b have the same sign. As a+b is positive, a and b are both positive. List all such integer pairs that give product 4.

$\sf \: 1,4 \\ \sf 2,2$

Calculate the sum for each pair.

$\sf \: 1+4=5 \\ \sf \: 2+2=4$

The solution is the pair that gives sum 4.

$\sf \: a=2 \\ \sf b=2$

Rewrite $\sf4x^{2}+4x+1 as \left(4x^{2}+2x\right)+\left(2x+1\right)$ .

$\sf \: \left(4x^{2}+2x\right)+\left(2x+1\right)$

Factor out 2x in 4x² + 2x.

$\sf \: 2x\left(2x+1\right)+2x+1$

Factor out common term 2x+1 by using distributive property.

$\sf\left(2x+1\right)\left(2x+1\right)$

Rewrite as a binomial square.

$\boxed{ \boxed{\bf\left(2x+1\right)^{2}} }$

__________________

<h3><u>3. 5n² + 10n + 20</u></h3>

$\sf \: 5 n ^ { 2 } + 10 n + 20$

Factor out 5. Polynomial n² + 2n+4 is not factored as it does not have any rational roots.

$\boxed{\boxed{\bf5\left(n^{2}+2n+4\right)} }$

<u>___________</u>_______

<h3><u>4. m³ - 729</u></h3>

$\sf \: m ^ { 3 } - 729$

Rewrite m³-729 as m³ - 9³. The difference of cubes can be factored using the rule: $a^{3}-b^{3}=\left(a-b\right)\left(a^{2}+ab+b^{2}\right)$ . Polynomial m²+9m+81 is not factored as it does not have any rational roots.

$\boxed{ \boxed{ \bf \: \left(m-9\right)\left(m^{2}+9m+81\right) }}$

__________________

<h3><u>5. x² - 81</u></h3>

$\sf \: x ^ { 2 } - 81$

Rewrite x²-81 as x² - 9². The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$ .

$\boxed{ \boxed{ \bf\left(x-9\right)\left(x+9\right) }}$

__________________

<h3><u>6. 15x² - 17x - 4</u></h3>

$\sf \: 15 x ^ { 2 } - 17 x - 4$

Factor the expression by grouping. First, the expression needs to be rewritten as 15x²+ax+bx-4. To find a and b, set up a system to be solved.

$\sf \: a+b=-17 \\ \sf \: ab=15\left(-4\right)=-60$

As ab is negative, a and b have the opposite signs. As a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -60.

$\sf \: 1,-60 \\ \sf 2,-30 \\ \sf3,-20 \\ \sf4,-15 \\ \sf5,-12 \\ \sf 6,-10$

Calculate the sum for each pair.

$\sf \: 1-60=-59 \\ \sf 2-30=-28 \\ \sf3-20=-17 \\ \sf4-15=-11 \\ \sf5-12=-7 \\ \sf6-10=-4$

The solution is the pair that gives sum -17.

$\sf \: a=-20 \\ \sf \: b=3$

Rewrite $\sf15x^{2}-17x-4$ as $\sf \left(15x^{2}-20x\right)+\left(3x-4\right)$ .

$\sf\left(15x^{2}-20x\right)+\left(3x-4\right)$

Factor out 5x in 15x²-20x.

$\sf \: 5x\left(3x-4\right)+3x-4$

Factor out common term 3x-4 by using distributive property.

$\boxed{\boxed{ \bf\left(3x-4\right)\left(5x+1\right) }}$

6 0

2 years ago

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