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ser-zykov [4K]
3 years ago
5

I running track is 400 meters. if i make 4 laps, what is the distance that i traveled?

Mathematics
1 answer:
IgorLugansk [536]3 years ago
5 0
You would have ran 1,200 meters
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HEEEEEEEEEEEEEEEEEEEEEEEEELLLLLLLLLLLLLLLLLLLLLLLPPPPPPPPPPPPPPP
zheka24 [161]
It is the third bubble
Hope i helped!
7 0
3 years ago
Read 2 more answers
10. A manufacturer wanted to know if more coupons would be redeemed if they were mailed to the
Mandarinka [93]

Using the t-distribution, it is found that the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

<h3>What are the hypothesis tested?</h3>

At the null hypothesis, it is tested if there is no difference, that is, the mean is of 0, hence:

H_0: \mu = 0

At the alternative hypothesis, it is tested if the mean number is greater for females, that is, the mean is greater than 0, hence:

H_1: \mu > 0

<h3>What is the test statistic?</h3>

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

In this problem, the parameters are given as follows:

\overline{x} = 1.5, \mu = 0, s = 4.75, n = 50.

Hence, the test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{1.5 - 0}{\frac{4.75}{\sqrt{50}}}

t = 2.23

<h3>What is the conclusion?</h3>

Considering a <em>right-tailed test</em>, as we are testing if the mean is greater than a value, with a <em>significance level of 0.01 and 50 - 1 = 49 df</em>, the critical value is given by t^{\ast} = 2.4.

Since the test statistic is less than the critical value for the right-tailed test, the data does not provide convincing evidence that the mean number is greater when the coupons were addressed to a female.

More can be learned about the t-distribution at brainly.com/question/26454209

5 0
2 years ago
The condition_______?proves that ∆ABC and ∆EFG are congruent by the SAS criterion.
snow_lady [41]

Answer:

(1)  D.Angle C is congruent to to Angle F. (2) C. SSS. (3) C. cannot be congruent to.

Step-by-step explanation:

1)

From the given figure it is noticed that

AC=EG

CB=GF

According to SAS postulate,  if two sides and the included angle of one triangle are congruent to two sides and the included angle of another triangle, then both triangles are congruent.

The included angles of congruent sides are angle C and angle G.

So, condition "Angle C is congruent to to Angle F"  will prove that the ∆ABC and ∆EFG are congruent by the SAS criterion.

2)

If AB\neq EF

According to SSS postulate,  if all three sides in one triangle are congruent to the corresponding sides in the other.

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore SSS criterion for congruence is violated.

3)

Since two corresponding sides are congruent but third sides of triangles are not congruent, therefore the included angle of congruent sides are different.

\angle C\neq \angle G

Therefore angle C and angle F cannot be congruent to each other.

4 0
3 years ago
What’s the vertex form of F(x)=x^2+2x-3
kupik [55]

Answer:

\large\boxed{f(x)=(x+1)^2-4}

Step-by-step explanation:

\text{The vertex form of a quadratic equation}\ f(x)=ax^2+bx+c:\\\\f(x)=a(x-h)^2+k\\\\(h,\ k)-\text{vertex}\\=====================================

\bold{METHOD\ 1:}\\\\\text{convert to the perfect square}\ (a+b)^2=a^2+2ab+b^2\qquad(*)\\\\f(x)=x^2+2x-3=\underbrace{x^2+2(x)(1)+1^2}_{(*)}-1^2-3\\\\f(x)=(x+1)^2-4\\==============================

\bold{METHOD\ 2:}\\\\\text{Use the formulas:}\ h=\dfrac{-b}{2a},\ k=f(k)\\\\f(x)=x^2+2x-3\to a=1,\ b=2,\ c=-3\\\\h=\dfrac{-2}{2(1)}=\dfrac{-2}{2}=-1\\\\k=f(-1)=(-1)^2+2(-1)-3=1-2-3=-4\\\\f(x)=(x-(-1))^2-4=(x+1)^2-4

4 0
4 years ago
Select the correct answer.
Neporo4naja [7]

Answer:

c salary refers to the fixed amount paid on a weekly bi-weekly

5 0
3 years ago
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