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2 years ago

Help with this and I’ll mark you as brainliest ❤️

Mathematics

1 answer:

Dahasolnce [82]2 years ago

8 0

Answer:

The answer is C

Step-by-step explanation:

Because 1 is 90 because it shaped like a L 2 is 45 because its narow, and 3 is the smallest of them all. I also did it before

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PLZZ HELP MEHH Sara graphs each quadratic equation. Which graph is the narrowest?

WINSTONCH [101]

Answer:

(B) y=2x^2

Step-by-step explanation:

(1/2) and (1/4) make the graph wider

2 makes it narrower

4 0

3 years ago

A manufacturer of paper coffee cups would like to estimate the proportion of cups that are defective (tears, broken seems, etc.)

Maksim231197 [3]

Answer:

The 95% confidence interval is $0.0503 < p < 0.1297$

The sample proportion is $\r p = 0.09$

The critical value is $Z_{\frac{\alpha }{2} } = 1.96$

The standard error is $SE =0.020$

Step-by-step explanation:

From the question we are told that

The sample size is n = 200

The number of defective is k = 18

The null hypothesis is $H_o : p = 0.08$

The alternative hypothesis is $H_a : p > 0.08$

Generally the sample proportion is mathematically evaluated as

$\r p = \frac{18}{200}$

$\r p = 0.09$

Given that the confidence level is 95% then the level of significance is mathematically evaluated as

$\alpha = 100 - 95$

$\alpha = 5\%$

$\alpha = 0.05$

Next we obtain the critical value of $\frac{ \alpha }{2}$ from the normal distribution table, the value is

$Z_{\frac{\alpha }{2} } = 1.96$

Generally the standard of error is mathematically represented as

$SE = \sqrt{\frac{\r p (1 - \r p)}{n} }$

substituting values

$SE = \sqrt{\frac{0.09 (1 - 0.09)}{200} }$

$SE =0.020$

The margin of error is

$E = Z_{\frac{ \alpha }{2} } * SE$

=> $E = 1.96 * 0.020$

=> $E = 0.0397$

The 95% confidence interval is mathematically represented as

$\r p - E < \mu < p < \r p + E$

=> $0.09 - 0.0397 < \mu < p < 0.09 + 0.0397$

=> $0.0503 < p < 0.1297$

7 0

3 years ago

(x^2−1)(3x−4)(3x+4) what is the product

Andrei [34K]

Answer:

9x^4-25x^2+16

Step-by-step explanation:

(x^2−1)(3x−4)(3x+4)=(*)

(x^2−1)((3x)^2−4^2) =

(x^2 - 1)(9x^2-16)=

x^2*9x^2 - x^2*16 - 1*9x^2 +16=

9x^4-16x^2-9x^2+16=

9x^4-25x^2+16

(*) (A-B)(A+B) =A^2 - B^2

6 0

2 years ago

What is the answer to 18 divided by parentheses 5-2 parentheses

deff fn [24]

18÷(5-2)

18÷3

6 is the answer

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