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3 years ago

Austin burns 120 calories per 30 minutes of weight training. How many calories does he burn in 90 minutes?

Mathematics

2 answers:

nadezda [96]3 years ago

6 0

Answer:

360 calories

Step-by-step explanation:

360

Fofino [41]3 years ago

5 0

Answer:

360

Step-by-step explanation:

You might be interested in

Tan (tt/2) =. i neeed help baddddd

Effectus [21]

Answer:

D: Undefined

Step-by-step explanation:

You can use the unit circle or your calc for this (don't forget to put it in radians!). I will use the unit circle to explain.

Tangent on the unit circle is on the xy lines themselves. tan (π/2) would be on the positive y axis. Remember that tan∅ is <em>sin∅/cos∅</em> and that the unit circle's radius is always one. If that is the case, the coordinate for tan (π/2) would be (0, 1). If cos is always x and sin is always y, you plug it in for tan∅. The fraction 1/0 is undefined, since you can't divide anything by 0. Therefore, tan (π/2) is undefined.

4 0

4 years ago

We'll assume that the stem diameter is normally distributed. An agronomist measured stem diameter in 8 randomly selected plants

BaLLatris [955]

Answer:

The 80% confidence interval for the population mean is (2.156, 2.394).

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 8 - 1 = 7

80% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 7 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.8}{2} = 0.9$ . So we have T = 1.415

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.415\frac{0.238}{\sqrt{8}} = 0.119$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is 2.275 - 0.119 = 2.156

The upper end of the interval is the sample mean added to M. So it is 2.275 + 0.119 = 2.394

The 80% confidence interval for the population mean is (2.156, 2.394).

6 0

3 years ago

What are the solutions to the equations?

Ghella [55]

Answer:

Step-by-step explanation:

x=(-4(plus minus)sqrt(4^2-4*2*(-1))/2*2

x=(-4(plus minus)sqrt(16+8))/4

x=(-4(plus minus)sqrt(24))/4

x=(-4(plus minus)sqrt(2*2*6))/4

x=(-4(plus minus)2*sqrt(6))/4

x=(2(-2(plus minus)*sqrt(6)))/2*2

x=(-2(plus minus)*sqrt(6))/2

x=(-2+sqrt(6))/2

x=(-2-sqrt(6))/2

(Option 4 I think, but can see for sure)

3 0

3 years ago

Prove :

Sauron [17]

Answer:

See Below.

Step-by-step explanation:

We want to verify the equation:

$\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

We can convert sec(α) to 1 / cos(α):

$\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Multiply both layers of the first fraction by cos(α):

$\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Simplify:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Subtract:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Distribute:

$\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Rewrite:

$\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Split:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{(1-\cos\alpha)(1+\cos\alpha)}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$