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2 years ago

9

Help pls I got an email from school Bc I didn’t finish this and I need to finish it before next class and I have like 6 more che

cks to do after eeeeeee!!!!!

1 answer:

denpristay [2]2 years ago

7 0

Answer:

169.25 $inches^{2}$

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Can someone help me with this problem please? I'll mark you as brainliest!

leonid [27]

Answer:

put a closed for on -1 and go to the left with an arrow

3 0

3 years ago

Read 2 more answers

If the polynomial

erik [133]

Answer:

(C) - x

Step-by-step explanation:

<u>Given polynomial:</u>

x¹⁹ + x¹⁷ + x¹³ + x⁷ + x⁵ + x³

<u>Group as follows:</u>

(x¹⁹ + x¹⁷) + (x¹³ + x¹¹) + (x⁷ + x⁵) + (x³ + x) - x =
x¹⁷(x² + 1) + x¹¹(x² + 1) + x¹⁵(x² + 1) + x(x² + 1) - x

As we see all terms have (x² + 1) as factor apart from the last one.

It means the remainder is - x

Correct choice is C

3 0

3 years ago

Here is the next one jaded :))

Ksju [112]

Answer:

x = 5.56

Step-by-step explanation:

Sin(17) = $\frac{x}{19}$

=> Isolate the "x"

Sin(17) * 19 = x

5.56 = x

Hope this helps!

6 0

3 years ago

From experience, an airline knows that only 80% of the passengers booked for a certain flight actually show up. If 9 passengers

Olenka [21]

Answer: 0.74

Step-by-step explanation:

Given : The probability that the passengers booked for a certain flight actually show up : p=0.80

The number of passengers are randomly selected : n= 9

Binomial probability formula to find the probability of getting success in x trial :-

$P(x)=^nC_xp^x(1-p)^{n-x}$

Then , the probability that more than 6 of them show up will be :-

$P(x>6)=P(7)+P(8)+P(9)\\\\=^9C_7(0.8)^7(0.2)^{2}+^9C_8(0.8)^8(0.2)^{1}+^9C_9(0.8)^9(0.2)^{0}\\\\=\dfrac{9!}{7!(9-7)!}(0.8)^7(0.2)^{2}+(9)(0.8)^8(0.2)+(1)(0.8)^9\approx0.3020+0.3020+0.1342=0.7382\approx0.74$

Hence, the probability that more than 6 of them show up = 0.74

7 0

3 years ago

Does this graph represent a function? Why or why not?

malfutka [58]

Answer:

Yes.

Step-by-step explanation:

It is a function because it passes the vertical line test. Each x-value has just one y-value. If the relation had any x-values that had several y-values, it would just be a relation. But this one is a function.

Hope this helps!

6 0

3 years ago