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Elena-2011 [213]
2 years ago
13

How to find the least value of quadratic expression 5-(x+3)^2

Mathematics
1 answer:
Mariana [72]2 years ago
5 0
Least value  is negative infinity
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Advocard [28]

Answer:

y = 38 + 72t - 16t^2

Step-by-step explanation:

So, solve for t when y=48.

The max height is of course, when t = 64/32 = 2.

5 0
2 years ago
Solve the equation pls <br> 4t=32
iren [92.7K]
The answer is 6. Because 4x6=32
8 0
2 years ago
Read 2 more answers
Pls Help me out homies ​
Alexandra [31]

Answer:

Just taking my points back

3 0
3 years ago
out of 30 questions on the maths test tomi knew the answer to 27 of them what percentage of questions did tomi know the answer f
Vedmedyk [2.9K]

Answer:

90

Step-by-step explanation:

\frac{27}{30}  \times 100

so after we cut the nessecary we get 90%

Hope this helps you

Mark as the brainlist

thank you

8 0
2 years ago
The heights of a random sample of 50 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centime
Minchanka [31]

Answer:

Step-by-step explanation:

Hello!

For me, the first step to any statistics exercise is to determine what is the variable of interest and it's distribution.

In this example the variable is:

X: height of a college student. (cm)

There is no information about the variable distribution. To estimate the population mean you need a variable with at least a normal distribution since the mean is a parameter of it.

The option you have is to apply the Central Limit Theorem.

The central limit theorem states that if you have a population with probability function f(X;μ,δ²) from which a random sample of size n is selected. Then the distribution of the sample mean tends to the normal distribution with mean μ and variance δ²/n when the sample size tends to infinity.

As a rule, a sample of size greater than or equal to 30 is considered sufficient to apply the theorem and use the approximation.

The sample size in this exercise is n=50 so we can apply the theorem and approximate the distribution of the sample mean to normal:

X[bar]~~N(μ;σ2/n)

Thanks to this approximation you can use an approximation of the standard normal to calculate the confidence interval:

98% CI

1 - α: 0.98

⇒α: 0.02

α/2: 0.01

Z_{1-\alpha /2}= Z_{1-0.01}= Z_{0.99} =2.334

X[bar] ± Z_{1-\alpha /2} * \frac{S}{\sqrt{n} }

174.5 ± 2.334* \frac{6.9}{\sqrt{50} }

[172.22; 176.78]

With a confidence level of 98%, you'd expect that the true average height of college students will be contained in the interval [172.22; 176.78].

I hope it helps!

4 0
2 years ago
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