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3 years ago

Please help a brotha out so I can get ungrounded! Will give Brainlyest!

Mathematics

1 answer:

kodGreya [7K]3 years ago

4 0

Answer:

38 degrees

Step-by-step explanation:

Angle ABE is 90 degrees, and angle FBE is 52 degrees. 90-52 is 38.

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Re-arrange this polynomial so its terms

Lera25 [3.4K]

Answer:

It should be 2x3 - 7x2 +5x - 15

Step-by-step explanation:

You have to look at the variable with the greatest exponential value

6 0

3 years ago

The volume of a cube is 1 cubic meter. What is the length of of each side of the cube

julsineya [31]

1 Metre is the length of each side of the cube.

To example, here is the Volume Of Cube Equation:

V = a^3

We are given the value for V, so just take the cube root of 1, which equates to 1.

4 0

3 years ago

A baseball player hits a ball at an angle of 52 degrees and at a height of 4.5 ft. if the ball's initial velocity after being hi

Dafna1 [17]

Answer:

$t\approx 5.865\,s$

Step-by-step explanation:

The motion equations that describe the ball are, respectively:

$x = \left[\left(152\,\frac{ft}{s} \right)\cdot \cos 52^{\circ} \right] \cdot t$

$y = 4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{ft}{s^{2}} \right) \cdot t^{2}$

The time required for the ball to hit the ground is computed from the second equation. That is to say:

$4.5\,ft + \left[\left(152\,\frac{ft}{s} \right)\cdot \sin 52^{\circ} \right] \cdot t - \frac{1}{2}\cdot \left(32.174\,\frac{m}{s^{2}} \right) \cdot t^{2} = 0$

Given that formula is a second-order polynomial, the roots of the equation are described below:

$t_{1}\approx 5.865\,s$ and $t_{2} \approx -0.048\,s$

Just the first root offers a realistic solution. Then, $t\approx 5.865\,s$ .

5 0

3 years ago

In an effort to estimate the mean amount spent per customer for dinner at a major Atlanta restaurant, data were collected for a

Softa [21]

Answer:

95% Confidence interval = (23.4,26.2)

Step-by-step explanation:

In this problem we have to develop a 95% CI for the mean.

The sample size is n=49, the mean of the sample is M=24.8 and the standard deviation of the population is σ=5.

We know that for a 95% CI, the z-value is 1.96.

The CI is

$M-z*\sigma/\sqrt{n}\leq\mu\leq M+z*\sigma/\sqrt{n}\\\\24.8-1.96*5/\sqrt{49}\leq\mu\leq 24.8+1.96*5/\sqrt{49}\\\\ 24.8-1.4\leq\mu\leq 24.8+1.4\\\\23.4\leq\mu\leq 26.2$

5 0

3 years ago

These box plots show daily low temperatures for a sample of days in two

9966 [12]

Answer:

B. The interquartile range (IQR) for town A, 20, is greater than the

IQR for town B, 10.

Step-by-step explanation:

IQR = Q₃ - Q₁

For the town A:

Q₃ = 40 and Q₁ = 20

IQR of town A = 40 - 20 = 20

For the town B:

Q₃ = 30 and Q₁ = 20

IQR of town A = 30 - 20 = 10

Check the given options:

The most statement is appropriate comparison of the spreads is B

B. The interquartile range (IQR) for town A, 20, is greater than the

IQR for town B, 10.

6 0

3 years ago