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1 year ago

15

Two friends, Damian and Nicole, took summer jobs. The equation y = 15.52

1 answer:

Sindrei [870]1 year ago

5 0

Answer:

sorry

Step-by-step explanation:

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A volleyball reaches its maximum of 13 feet, 3 seconds after it is served. What quadratic formula could model the height of the

Katena32 [7]

Answer:

$H=-(t-3)^2+13$

Step-by-step explanation:

The path covered by the volleyball will be a downward parabola with the vertex being the highest point of the ball.

A general form of a downward parabola is given as:

$y=-a(x-h)^2+k$

Where $(h,k)$ is the vertex of the parabola and 'a' is a constant.

Now, let 'h' be the vertical height and 't' be the time taken.

So, the equation would be of the form:

$H=-a(t-h)^2+k$

Now, as per question:

h = 2 seconds, k = 13 feet.

$H=-a(t-3)^2+13$

Now, taking a = 1. So, the formula that can be used is:

$H=-(t-3)^2+13$

7 0

2 years ago

I need help ,, i don't understand !<br>

otez555 [7]

Answer: B

Step-by-step explanation: The question is asking for the point greatly different from all the rest, the outlier. Point B is far away from the grouping of all the other points, making it the outlier.

7 0

2 years ago

ILL GIVE U STUFFFFF PLZZZZZZZZZZZ ANSWER ITTTTTTTTT

belka [17]

Answer:

Slope: $-\frac{3}{4}$

Y-intercept: 399

Slope-intercept form: $y=-\frac{3}{4}x+399$

Step-by-step explanation:

1. The equation in Standard form is given in the problem:

$3y+4y=1596$

2. Then, to write the equation of the line in slope-intercept form $y=mx+b$ , where <em>m</em> is the slope and <em>b</em> is the y-intercept, you must solve for <em>y</em> as you can see below:

$4y=-3x+1596\\y=\frac{-3x+1596}{4}\\y=-\frac{3}{4}x+399$

3. Therefore, the slope and the y-intercept are:

$m=-\frac{3}{4}\\b=399$

5 0

2 years ago

(a) Solve the following formula for b.<br>A= 1÷2ab please!!!<br>

Anvisha [2.4K]

Answer:

2

Step-by-step explanation:

a = 1/2ab

~Divide 1/2a to both sides

2 = b

Best of Luck!

3 0

2 years ago

Read 2 more answers

Suppose that the mean water hardness of lakes in Kansas is 425 mg/L and these values tend to follow a normal distribution. A lim

tino4ka555 [31]

Answer:

The mean water hardness of lakes in Kansas is 425 mg/L or greater.

Step-by-step explanation:

We are given the following data set:

346, 496, 352, 378, 315, 420, 485, 446, 479, 422, 494, 289, 436, 516, 615, 491, 360, 385, 500, 558, 381, 303, 434, 562, 496

Formula:

$\text{Standard Deviation} = \sqrt{\displaystyle\frac{\sum (x_i -\bar{x})^2}{n-1}}$

where $x_i$ are data points, $\bar{x}$ is the mean and n is the number of observations.

$Mean = \displaystyle\frac{\text{Sum of all observations}}{\text{Total number of observation}}$

$Mean =\displaystyle\frac{10959}{25} =438.36$

Sum of squares of differences = 175413.76

$S.D = \sqrt{\dfrac{175413.76}{24}} = 85.49$

Population mean, μ = 425 mg/L

Sample mean, $\bar{x}$ = 438.36

Sample size, n = 25

Alpha, α = 0.05

Sample standard deviation, s = 85.49

First, we design the null and the alternate hypothesis

$H_{0}: \mu \geq 425\text{ mg per Litre}\\H_A: \mu < 425\text{ mg per Litre}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{438.36 - 425}{\frac{85.49}{\sqrt{25}} } = 0.7813$

Now, $t_{critical} \text{ at 0.05 level of significance, 24 degree of freedom } = -1.7108$

Since,

The calculated t-statistic is greater than the critical value, we fail to reject the null hypothesis and accept it.

Thus, the mean water hardness of lakes in Kansas is 425 mg/L or greater.

6 0

2 years ago