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Rainbow [258]
3 years ago
13

The diameter of a circle is 4 inches. What is the circle's area? d=4 in Use 3.14 for u.

Mathematics
1 answer:
harina [27]3 years ago
7 0

Answer:

The area would be 12.57

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What's 2/3 divided by 1/4
svlad2 [7]

Answer:

8/3 or 2 and 2/3

Step-by-step explanation:

5 0
3 years ago
Read 2 more answers
Help if you can do this please
olga nikolaevna [1]

Answer:

Step-by-step explanation:

a. Since the parabola is compressed by a factor of 1/3 we can state:

  • a parabola is written this way : y=(x-h)²+k
  • h stands for the translation to the left ⇒ 2*3=6
  • k for the units down  ⇒4*3=12

So the equation is : y=(x-6)²+12

b.Here the parabola is stretched by a factor of 2 so we must multiply by 1/2

  • We khow that a parabola is written this way : y=(x-h)²+k
  • (h,k) are the coordinates of the vertex
  • the maximum value is  7*0.5=3.5
  • we khow tha the derivative of a quadratic function is null in the maximum value
  • so let's derivate (x-h)²+k= x²+h²-2xh+k
  • f'(x)= 2x-2h    h is 1 since the axe of simmetry is x=1
  • f'(x)=2x-2 ⇒2x-2=0⇒ x= 1
  • Now we khow that 1 is the point where the derivative is null
  • f(1)=3.5
  • 3.5=(x-1)²+k
  • 3.5= (1-1)²+k⇒ k=3.5

So the equation is : y=(x-1)²+3.5

7.

the maximum height is where the derivative equals 0

  • h= -5.25(t-4)²+86
  • h= -5.25(t²-8t+16)+86
  • h=-5.25t²+42t-84+86
  • h=-5.25t²+42t+2

Let's derivate it :

  • f(x)= -10.5t+42
  • -10.5t+42=0
  • 42=10.5t
  • t= 42/10.5=4

When the height was at max t=4s

  • h(max)= -5.25(4-4)²+86 = 86 m

h was 86m

8 0
2 years ago
Complete the square to solve<br> LaTeX: 0=-x^2+4x-5<br> Also written as<br> 0= -x^2+4x-5
Feliz [49]
X=2+-(plus or minus)i
7 0
3 years ago
Your friend is graphing the point (5, 3). His first step is to start at the origin and go up 5 spaces. Which statement about his
ddd [48]
He is going up 5 spaces.
6 0
3 years ago
The attention span of children (ages 3 to 5) is claimed to be Normally distributed with a mean of 15 minutes and a standard devi
Agata [3.3K]

Answer:

If the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 15 minutes

Sample size, n = 10

Alpha, α = 0.05

Population standard deviation, σ = 4 minutes

First, we design the null and the alternate hypothesis

H_{0}: \mu = 15\text{ minutes}\\H_A: \mu > 15\text{ minutes}

Since the population standard deviation is given, we use one-tailed z test to perform this hypothesis.

Formula:

z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }

Now, z_{critical} \text{ at 0.05 level of significance for one tail} = 1.64

Thus, we would reject the null hypothesis if the z-statistic is greater than this critical value.

Thus, we can write:

z_{stat} = \displaystyle\frac{\bar{x} - 15}{\frac{4}{\sqrt{10}} } > 1.64\\\\\bar{x} - 15>1.64\times \frac{4}{\sqrt{10}}\\\bar{x} - 15>2.07\\\bar{x} > 15+2.07\\\bar{x} > 17.07

Thus, the decision rule would be if the observed sample mean is greater than 17.07 minutes, then we would reject the null hypothesis.

6 0
3 years ago
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