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3 years ago

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Which terms accurately classify this triangle? Choose exactly two answers that are correct. A. scalene B. obtuse C. acute D. iso

sceles

1 answer:

ser-zykov [4K]3 years ago

8 0

The answer is isosceles and scalene.

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Which option summarizes what gave Chandra Gupta I power

mars1129 [50]

Answer:

Chandra Gupta I, King of India reigned from 320 to about 330 CE and founder of the Gupta empire. He was the grandson of Sri Gupta, the first known leader of the Gupta line. Chandra Gupta I, whose first life is unknown, became the local chief of the kingdom of Magadha parts of the modern state of Bihar.

Step-by-step explanation:

6 0

3 years ago

A survey asked people to name their favorite season. The results of the survey showed that 34% said spring and 15% said fall. Wh

max2010maxim [7]

Answer:

D. 29% and 22%

Step-by-step explanation:

The known percentages are 34% and 15%. Add these two together to get 49%. All four choices should add up to 100%. So subtract to find what remains:

100% - 49%

= 51%

We don't actually have enough information to determine the percents for summer and winter. BUT the only answer that adds up to 51% is answer D. 29% and 22%.

8 0

2 years ago

The polynomial x3+8 is equal to

s344n2d4d5 [400]

The polynomial x3+8 is equal to (x+2)(x2-2x+4)

3 0

3 years ago

Read 2 more answers

Consider three events A, B and C with following properties.

lisov135 [29]

By the inclusion/exclusion principle,

$\mathbb P(D)=\mathbb P(A\cup B\cup C)$
$\mathbb P(D)=\mathbb P(A)+\mathbb P(B)+\mathbb P(C)-\bigg(\mathbb P(A\cap B)+\mathbb P(A\cap C)+\mathbb P(B\cap C)\bigg)+\mathbb P(A\cap B\cap C)$
$\mathbb P(D)=\dfrac14+\dfrac16+\dfrac14-\dfrac39+\dfrac1{13}$
$\mathbb P(D)=\dfrac{16}{39}\neq1$

So the union of A, B, and C does not constitute the entire sample space.

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

3 years ago