Does the graph represent a function?

Passing through (1,9) parallel to y=5x+6

AysviL [449]

Gradient = 5

9 = 5 + c
C = 4

Answer:
Y=5x+4

5 0

3 years ago

There are red and blue balls in a bag. The number of blue balls is 27 and the red balls are 4/ 7 of all balls. What is the total

aliina [53]

<h3>♫ - - - - - - - - - - - - - - - ~Hello There!~ - - - - - - - - - - - - - - - ♫</h3>

➷ We know this:

3/7 = 27

You need to find the value of 1/7

To do this, divide by 3

27/3 = 9

1/7 = 9

Multiply this by 7 to get the total number of balls:

9 x 7 = 63

There are 63 balls.

➶ Hope This Helps You!

➶ Good Luck (:

➶ Have A Great Day ^-^

↬ ʜᴀɴɴᴀʜ ♡

7 0

3 years ago

Write an equation perpendicular to x-4y= 20 that passes through the point (2, -5)

KIM [24]

4(x -2) +(y +5) = 0 . . . . . an equation

4x +y = 3 . . . . . . . . . . . . the equation in standard form
___
A graph can be seen at
https://www.desmos.com/calculator/3ct9gfxclj

5 0

3 years ago

Tennis elbow is thought to be aggravated by the impact experienced when hitting the ball. The article "Forces on the Hand in the

Gnoma [55]

Answer:

Step-by-step explanation:

Hello!

The objective is to study whether there is a greater force after impacting on one- handed backhand drive in advanced tennis players than in intermediate tennis players.

Sample 1: Advanced tennis players

X₁: Force (N) on the hand just after impact on a one- handed backhand drive for an advanced tennis player.

n₁= 6

X[bar]₁= 40.29 N

S₁= 11.29

Sample 2: Intermediate players

X₂: Force (N) on the hand just after impact on a one- handed backhand drive for an intermediate tennis player.

n₂= 8

X[bar]₂= 21.40

S₂= 8.30

Assuming that both variables have a normal distribution and both population variances are equal, to compare these two populations is best to do so trough their population means using a t-test for independent samples.

If the force is greater for the advanced players than for the intermediate players, then you'd expect the population mean for the advanced players to be greater than the population mean for the intermediate players:

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

α: 0.05

$t= \frac{(X_[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } } ~~t_{n_1+n_2-2}$

$Sa= \sqrt{\frac{(n_1-1)S_1^2+(n_2-1)S_2^2}{n_1+n_2-2} } = \sqrt{\frac{5*127.51+7*68.92}{6+8-2} }= 9.66$

$t_{H_0}= \frac{(40.29-21.40)-0}{9.66\sqrt{\frac{1}{6} +\frac{1}{8} } } = 3.62$

Using the p-value approach, the decision rule is

If p-value ≤ α, reject the null hypothesis

If p-value > α, do not reject the null hypothesis

The p-value for this test is 0.00024, it is less than the level of significance, so the decision is to reject the null hypothesis.

This means that at a 5% significance level you can conclude that the average force experienced on the hand after a one-handed backhand drive for advanced players is greater than the average force experienced on the hand after a one-handed backhand drive for intermediate players.

I hope this helps!

3 0

4 years ago

Consider two independent tosses of a fair coin. Let A be the event that the first toss results in heads, let B be the event that

aliina [53]

Answer with Step-by-step explanation:

We are given that two independent tosses of a fair coin.

Sample space={HH,HT,TH,TT}

We have to find that A, B and C are pairwise independent.

According to question

A={HH,HT}

B={HH,TH}