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denis23 [38]
3 years ago
8

Can someone please help with this problem

Mathematics
1 answer:
JulsSmile [24]3 years ago
8 0

Answer:

f(x)=-3(x+2)^2-2

Step-by-step explanation: the the FX graph the -3 in my equation represents the decrease(for being negative) in the graph, the positive 2 represents the x axis if that number is positive the negative 3 is going to change the x axis then will end up being negative which is why the x starts off in negative 2, the last number represents the y axis if the y axis is negative 2 the y will start off at -2

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If a fair-number cube with sides numbered from 1 to 6 is tossed 240 times, approximately how many times would the fair-number cu
LenKa [72]
It wouls be greater 50times
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3 years ago
20 as a percentage of 50
Y_Kistochka [10]
20:50 as 40:100
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3 years ago
Read 2 more answers
I toss an unfair coin 12 times. This coin is 65% likely to show up heads. Calculate the probability of the following.
mars1129 [50]

Answer:

a. 0.0368

b. 0.99992131

c. 0.2039

d. 0.0048

e. 0.6533

Step-by-step explanation:

Let the probability of obtaining a head be p = 65% = 13/20 = 0.65. The probability of not obtaining a head is q = 1 - p = 1 -13/20 = 7/20 = 0.35

Since this is a binomial probability, we use a binomial probability.

a. The probability of obtaining 11 heads is ¹²C₁₁p¹¹q¹ = 12 × (0.65)¹¹(0.35) = 0.0368

b. Probability of 2 or more heads P(x ≥ 2) is

P(x ≥ 2) = 1 - P(x ≤ 1)

Now P(x ≤ 1) = P(0) + P(1)

= ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹

= (0.65)⁰(0.35)¹² + 12(0.65)¹(0.35)¹¹

= 0.000003379 + 0.00007531

= 0.0007869

P(x ≥ 2) = 1 - P(x ≤ 1)

= 1 - 0.00007869

= 0.99992131

c. The probability of obtaining 7 heads is ¹²C₇p⁷q⁵ = 792(0.65)⁷(0.35)⁵ = 0.2039

d. The probability of obtaining 7 heads is ¹²C₉q⁹p³ = 220(0.65)³(0.35)⁹ = 0.0048

e. Probability of 8 heads or less P(x ≤ 8) = ¹²C₀p⁰q¹² + ¹²C₁p¹q¹¹ + ¹²C₂p²q¹⁰ + ¹²C₃p³q⁹ + ¹²C₄p⁴q⁸ + ¹²C₅p⁵q⁷ + ¹²C₆p⁶q⁶ + ¹²C₇p⁷q⁵ + ¹²C₈p⁸q⁴

= = ¹²C₀(0.65)⁰(0.35)¹² + ¹²C₁(0.65)¹(0.35)¹¹ + ¹²C₂(0.65)²(0.35)¹⁰ + ¹²C₃(0.65)³(0.35)⁹ + ¹²C₄(0.65)⁴(0.35)⁸ + ¹²C₅(0.65)⁵(0.35)⁷ + ¹²C₆(0.65)⁶(0.35)⁶ + ¹²C₇(0.65)⁷(0.35)⁵ + ¹²C₈(0.65)⁸(0.35)⁴

= 0.000003379 + 0.00007531 + 0.0007692 + 0.004762 + 0.01990 + 0.05912 + 0.1281 + 0.2039 + 0.2367

= 0.6533

3 0
3 years ago
For an angle 0 with the point (-20,-21) on its terminating side, what is the value of cosine ?
Karo-lina-s [1.5K]

check the picture below.


\bf (\stackrel{a}{-20},\stackrel{b}{-21})\impliedby \textit{let's find the \underline{hypotenuse}} \\\\\\ \textit{using the pythagorean theorem} \\\\ c^2=a^2+b^2\implies c=\sqrt{a^2+b^2} \qquad \begin{cases} c=hypotenuse\\ a=\stackrel{adjacent}{-20}\\ b=\stackrel{opposite}{-21}\\ \end{cases} \\\\\\ c=\sqrt{(-20)^2+(-21)^2}\implies c=\sqrt{841}\implies c=29 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill cos(\theta )=\cfrac{\stackrel{adjacent}{-20}}{\stackrel{hypotenuse}{29}}~\hfill

5 0
3 years ago
Which of the following statements is true about an image after a dilation?
tino4ka555 [31]

Answer:

Not sure what the answer choices are, but choose the choice that says the new image is either stretched or shrunk. In a dilation, the shape/corresponding sides of the pre-image are preserved in the new image, but the size of the new image is altered.

Step-by-step explanation:

hope this helps!

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