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3 years ago

Multiply the polynomial 10. (3-2v)(2-5v) 11. (5x-4)(3x2+x-5)

Mathematics

1 answer:

tino4ka555 [31]3 years ago

5 0

Answer:

10) $10v^2-19v+6$

11) $15x^3-7x^2-29x+20$

Step-by-step explanation:

Well first let's make the box.

10)

Look at the image below ↓

$10v^2-19v+6$

11)

Look at the image below ↓

$15x^3-7x^2-29x+20$

Thus,

number 10 is $10v^2-19v+6$ and number 11 is $15x^3-7x^2-29x+20$ .

Hope this helps :)

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The advertised duration of the trip is 2 hours and 15 minutes. The company operating the ferry would like to increase its operat

vesna_86 [32]

The 2.88 minutes have to reduce the standard deviation to increase the efficiency.

According to the statement

we have given that the duration of the trip is 2 hours and 15 minutes. And we have to how much standard deviation be reduced to increase the operational efficiency.

So, For this purpose, we know that the

Let X be the time taken to reach the resort. Then it is given that the

N (μ , σ).

Here μ is 2 hours then

μ = 2*60 = 120 minutes.

And

Let σ be the reduced standard deviation such that new distributions becomes N (μ , σ)

Then P ( x <135)

here 135 is the minute format of 2 hours and 15 min.

So,

P ( x <135) = 0.95

And

P [ (x -120 / σ) < (135 -120 / σ)] = 0.95

P [ Z < ( 15 / σ)] = 0.95

And after solving it the

σ = 9.22 min.

i.e. 12 - 9.22 = 2.88 min.

So, The 2.88 minutes have to reduce the standard deviation to increase the efficiency.

Learn more about Standard deviation here

brainly.com/question/475676

Disclaimer: This question was incomplete. Please find the full content below.

Question:

The time it takes for a ferry to reach a summer resort from the mainland is Normally distributed with mean 2 hours and standard deviation 12 minutes. The advertised duration of the trip is 2 hours and 15 minutes. The company operating the ferry would like to increase its operational efficiency by reducing the standard deviation so that they are on time 95% of the time. How much do they have to reduce the standard deviation

#SPJ4

5 0

1 year ago

A parallelogram is a _____. (Choose all that apply)

Mama L [17]

Answer:

Quadrilateral

Step-by-step explanation:

The definition of a parallelogram is a quadrilateral that has two sets of parallel sides.

3 0

2 years ago

Read 2 more answers

Normal probability Density Function Use a graphing utility to graph the model probability density function with µ = 0 and ð = 2,

Nuetrik [128]

Answer:

Step-by-step explanation:

Hello!

The standard deviation (δ) is a measure of variability, this means, it shows how dispersed the data set is with respect to the mean. The population mean (μ) is a measurement of position. The three graphics have the same position μ=0 but their standard deviations change, this means, the form of their bells is different. The greater the value of the standard deviation, the more dispersed the data is you can see this graphically because the width of the bell will be greater.

Graph attached.

I hope it helps!

5 0

3 years ago

Angela invests 2,550 at 3% interest compounded annually. What will be the balance in the account after 1.5 years?

marin [14]

answer my question then i'll answer yours

Which sentence uses a conjunctive adverb correctly?

The kite was hopelessly tangled in electrical wires, still, the little girl's father struggled to get it down.The kite was hopelessly tangled in electrical wires; still, the little girl's father struggled to get it down.

6 0

2 years ago

Write the equation of a hyperbola with vertices (0, -4) and (0, 4) and foci (0, -5) and (0, 5).

andre [41]

Check the picture below. So, more or less looks like so.

notice, the center is clearly at the origin, and notice how long the "a" component is, also, bear in mind that, is opening towards the y-axis, that means the fraction with the "y" variable is the positive one.

Also notice, the "c" distance from the center to either foci, is just 5 units.

$\bf \textit{hyperbolas, vertical traverse axis }\\\\
\cfrac{(y-{{ k}})^2}{{{ a}}^2}-\cfrac{(x-{{ h}})^2}{{{ b}}^2}=1
\qquad 
\begin{cases}
center\ ({{ h}},{{ k}})\\
vertices\ ({{ h}}, {{ k}}\pm a)\\
c=\textit{distance from}\\
\qquad \textit{center to foci}\\
\qquad \sqrt{{{ a }}^2+{{ b }}^2}
\end{cases}\\\\
-------------------------------\\\\$

$\bf \begin{cases}
h=0\\
k=0\\
a=4\\
c=5
\end{cases}\implies \cfrac{(y-{{ 0}})^2}{{{ 4}}^2}-\cfrac{(x-{{ 0}})^2}{{{ b}}^2}=1\implies \cfrac{y^2}{16}-\cfrac{x^2}{b^2}=1
\\\\\\
c=\sqrt{a^2+b^2}\implies c^2=a^2+b^2\implies \sqrt{c^2-a^2}=b
\\\\\\
\sqrt{5^2-4^2}=b\implies \boxed{3=b}
\\\\\\
\cfrac{y^2}{16}-\cfrac{x^2}{3^2}=1\implies \boxed{\cfrac{y^2}{16}-\cfrac{x^2}{9}=1}$

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3 years ago