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worty [1.4K]
3 years ago
10

The diagram on the right shows the graph of a quadratic function f(x)=kx^2+6x+h.Point A(3,14) is the maximum point of the graph

of a quadratic function.
(a)Given k is an integer where -2<k<2,state the value of k.​

Mathematics
1 answer:
weeeeeb [17]3 years ago
3 0

Answer:

4k

Step-by-step explanation:

4k

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1 True

2 c>−4

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6 0
3 years ago
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Name a rational number between the two given numbers? 7/11ths and 4/7ths explain how you know​
Alex_Xolod [135]

Answer:

basically u have to beble gep

Step-by-step explanation:

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4 0
3 years ago
A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% of all circu
telo118 [61]

Answer:

64.56% probability that between 17 and 25 circuits in the sample are defective.

Step-by-step explanation:

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

E(X) = np

The standard deviation of the binomial distribution is:

\sqrt{V(X)} = \sqrt{np(1-p)}

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that \mu = E(X), \sigma = \sqrt{V(X)}.

In this problem, we have that:

n = 110, p = 0.2

So

\mu = E(X) = np = 110*0.2 = 22

\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{110*0.2*0.8} = 4.1952

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

Z = \frac{X - \mu}{\sigma}

Z = \frac{25 - 22}{4.1952}

Z = 0.715

Z = 0.715 has a pvalue of 0.7626.

X = 17

Z = \frac{X - \mu}{\sigma}

Z = \frac{17 - 22}{4.1952}

Z = -1.19

Z = -1.19 has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.

4 0
3 years ago
Pls help i might give brainliest
zmey [24]

Answer:

60% I hope I help :) have a great day

6 0
2 years ago
2990000 divided by 56276
marta [7]

Answer:

I'm sorry, I think you mistyped, but the answer for this is 53.130997227947970715758049612623

Have a nice day! :)

-Vana

3 0
3 years ago
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