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1 year ago

15

HELP!!!

1 answer:

MakcuM [25]1 year ago

8 0

Answer:

based on my experience mind you're answer.

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Find a primitivism root of 6

avanturin [10]

Answer:

6, 5, 47, 5, 103, 5, 167, 5

8 0

2 years ago

Solve the equations X^2+y^2 =16 x+y=1

Leviafan [203]

Answer:

x = 1-y/6 and y= - 35/ - 37

Step-by-step explanation:

5 0

2 years ago

John is interested in purchasing a multi-office building containing five offices. The current owner provides the following proba

Elenna [48]

Answer:

Option B.

Step-by-step explanation:

The given table is:

Number of Lease Offices : 0 1 2 3 4 5

Probability : 5/18 1/4 1/9 1/18 2/9 1/12

The expected probability is

Expected probability = $\sum_{i=0}^5 x_{i}p(x_i)$

Expected probability = $0p(0)+1P(1)+2P(2)+3P(3)+4P(4)+5P(5)$

Expected probability = $0\cdot (\frac{5}{18})+1\cdot (\frac{1}{4})+2\cdot (\frac{1}{9})+3\cdot (\frac{1}{18})+4\cdot (\frac{2}{9})+5\cdot (\frac{1}{12})=\frac{35}{18}$

It is given that the yearly lease = $12,000.

The yearly leases for the whole building in a given year is

Yearly leases = $\frac{35}{18}\times 12000=23333.3333333\approx 23333.33$

Therefore, the correct option is B.

3 0

3 years ago

Coefficiants of (2x+y)^4

sattari [20]

By the binomial theorem,

$(2x+y)^4=\displaystyle\sum_{k=0}^4\binom 4k(2x)^{4-k}y^k=\sum_{k=0}^4\binom 4k2^{4-k}x^{4-k}y^k$

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

Then the coefficients of the $x^{4-k}y^k$ terms in the expansion are, in order from $k=0$ to $k=4$ ,

$\dbinom 402^{4-0}=1\cdot2^4=16$

$\dbinom412^{4-1}=4\cdot2^3=32$

$\dbinom422^{4-2}=6\cdot2^2=24$

$\dbinom432^{4-3}=4\cdot2^1=8$

$\dbinom442^{4-4}=1\cdot2^0=1$

3 0

3 years ago

Polygon PQRS, shown in the figure, is dilated by a scale factor of 0.5 with the origin as the center of dilation, resulting in p

nata0808 [166]

Based on the above:

The slope of P'Q' is = -3/2

The length of P'Q' is approximately = $\sqrt[3]{13}$

<h3>What is the polygon about?</h3>

The slope of P'Q'

= -6/4 = -3/2

The length of P'Q' =

P'Q' = $\sqrt{6^2 + 4^2}$

= $\sqrt{52}$

= $\sqrt[2]{13}$

Therefore;

P'Q' = $\frac{3}{2}$

P'Q' = $\frac{3}{2}$ x $\sqrt[2]{13}$

= $\sqrt[3]{13}$

Learn more about polygon from

brainly.com/question/1592456

#SPJ1

7 0

1 year ago