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3 years ago

The area of a square carpet tile is 900 square centimeters. What is the length of one edge of the tile in centimeters

1 answer:

8 0

Let the side of one edge of tile = a cm.

So area of tile = (side)^2 = $a^2$

But area of tiles = 900 square centimeter.

So $a^2 = 900$

$a = \sqrt{900}$
$= 30$

So side of one edge of tile is = 30 cm

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Find a so that the point (-1, 2) is on the graph of f(x)=ax^2+5.

Gnesinka [82]

We are asked to determine the value of a such that the function f(x) = ax^2 + 5 is fit for the point given (-1,2). In this case, we substitute 2 to y and -1 to x. The result is then 2 = a*(-1)^2 + 5 ; 2 = a + 5; a is then equal to -3

4 0

4 years ago

Tanh(2a)==1/2 then a=?

stepladder [879]

Answer:

2a=1/2

a=1/4

Step-by-step explanation:

6 0

3 years ago

Change 1/7 / 4/21 to a ratio?<br>

sergeinik [125]

Answer:

3:4

Step-by-step explanation:

(1/7)/(4/21) = (1/7)×(21/4) = 3/4 = 3:4

6 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

CAN SOMEONE HELP WITH THIS??????????????

svetoff [14.1K]

Answer:

8963.46

Step-by-step explanation:

84*66 for the rectangle and since there is two semi circles you could just calculate it as one so it would be 84*66+33*33*3.14 because formula for area of circle is A= pi*raduis with exponant of 2

3 0

3 years ago