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2 years ago

First to answer correctly will be marked brainliest 40 points as well

Mathematics

1 answer:

skelet666 [1.2K]2 years ago

7 0

Answer:

see explanation↓

Step-by-step explanation:

(a)

$(3x+2)-(x+5)=3x+2-x-5$

$(3x+2)-(x+5)=2x-3$

ANSWER: Linear binomial

(b)

$(3x+8)+(4x^{2}-x)=3x+8+4x^{2}-x$

$(3x+8)+(4x^{2}-x)=4x^{2} +2x+8$

ANSWER: Quadratic trinomial

(c)

$(x^{2}+7x)-(x^{2}+5)=x^{2}+7x-x^{2}-5$

$(x^2+7x)-(x^2+5)=7x-5$

ANSWER: Linear binomial

(d)

$(2x+8)+(3x^2-2)=2x+8+3x^2-2$

$(2x+8)+(3x^2-2)=3x^2+2x+6$

ANSWER: Quadratic trinomial

(e)

$(x^2+5x-2)+(8-5x)=x^2+5x-2+8-5x$

$(x^2+5x-2)+(8-5x)=x^2+6$

ANSWER: Quadratic binomial

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Step-by-step explanation:

To solve this subject of the formulae given that A = πab.

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3 years ago

spin [16.1K]

Answer:

AAS

Step-by-step explanation:

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3 years ago

Read 2 more answers

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!

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4 years ago