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TEA [102]
4 years ago
6

Quick help. Will Mark as Brainliest if my laptop stops acting up.

Mathematics
1 answer:
Dimas [21]4 years ago
8 0
Vertex is now at (-1,5)

for
y=a(x-h)^2+k
vertex is (h,k)
so veertex is (-1,5)

y=a(x-(-1))^2+5
y=a(x+1)^2+5
a is a constant, we will asssume that it is 1 because all the choices have 1

y=1(x+1)^2+5
y=(x+1)^2+5

2nd option
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What is 10/12 times 3/5 times 1/4 equal
natima [27]
1/8 is the answer jahdbdjdndndndn
3 0
3 years ago
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This semicircle has a diameter of 5 m.
Mumz [18]
3.93  is the correct answer.
I hope this helps.
8 0
4 years ago
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36 out of 100 randomly selected taxpayers knew about tax incentives for installing energy-saving furnaces. Find a 90% confidence
tresset_1 [31]

Answer:

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

For this problem, we have that:

n = 100, \pi = \frac{36}{60} = 0.6

90% confidence level

So \alpha = 0.1, z is the value of Z that has a pvalue of 1 - \frac{0.1}{2} = 0.95, so Z = 1.645.

The lower limit of this interval is:

\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 - 1.645\sqrt{\frac{0.36*0.64}{100}} = 0.28

The upper limit of this interval is:

\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.36 + 1.645\sqrt{\frac{0.36*0.64}{100}} = 0.44

The 90% confidence interval for the population proportion who knew about the incentives is (0.28, 0.44).

7 0
3 years ago
in how many ways can a student arrange 6 textbooks on a locker shelf that can hold 4 books at a time? 30 120 240 360
denis23 [38]
On the 1st place he can put 6 textbooks, on the 2nd 5 textbooks etc...
6 · 5 · 4 · 3 = 360
Answer: D) He can arrange textbooks in 360 ways.
4 0
3 years ago
40% of Oatypop cereal boxes contain a prize. Hannah plans to keep buying cereal until she gets a prize. What is the probability
aleksley [76]

Answer:

(0.6)^{n}+n(0.4)(0.6)^{n-1}+\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}

Step-by-step explanation:

We will use the binomial distribution. Let X be the random variable representing the no. of boxes Hannah buys before betting a prize.

Our success is winning the prize, p =40/100 = 0.4

Then failure q = 1-0.4  = 0.6

Hannah keeps buying cereal boxes until she gets a prize. Then n be no. times she buys the boxes.

P(X ≤ 3) = P(X=0) +P(X=1)+P(X=2)+P(X=3)

             = \binom{n}{0}p^{0}q^{n-0} +  \binom{n}{1}p^{1}q^{n-1}+ \binom{n}{2}p^{2}q^{n-2}+ \binom{n}{3}p^{3}q^{n-3}


             =  q^{n}+npq^{n-1}+\frac{n(n-1)(p)^{2}q^{n-2}}{2}+\frac{n(n-1)(n-2)p^{3}q^{n-3}}{6}

             = (0.6)^{n}+n(0.4)(0.6)^{n-1}+[tex]\frac{n(n-1)(0.4)^{2}0.6^{n-2}}{2} +\frac{n(n-1)(n-2)0.4^{3}0.6^{n-3}}{6}


7 0
3 years ago
Read 2 more answers
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