What is the measure of the obtuse angle in this figure?<br> 5* *4°/x01

Triangle ABC has side lengths 8 cm, 17 cm, and x cm. The side length of x is shorter than the longest side. What does the value

snow_tiger [21]

Answer:

8 cm and 15 cm

8 0

3 years ago

Parallel lines x and w are cut by transversals z and y and form 4 angles at each intersection. Where line x intersects with line

ale4655 [162]

Answer:

$m\angle 1 =105^\circ\\m\angle 2 =75^\circ$

Step-by-step explanation:

The diagram is drawn and attached below.

From the diagram

$m\angle 2 =75^\circ $ (Alternate Angles)$

$m\angle 1+ m\angle 2 =180^\circ $ (Sum of angles on a straight line)\\m\angle 1+75^\circ =180^\circ\\m\angle 1 =180^\circ-75^\circ\\m\angle 1 =105^\circ$

7 0

3 years ago

Solve for u<br> 3u = 78<br> Simplify your answer as much as possible.

scZoUnD [109]

You simply divide 78 by 3, so the answer will be 26.

8 0

3 years ago

Read 2 more answers

The average breaking strength of a certain brand of steel cable is 2000 pounds, with a standard deviation of 100 pounds. A sampl

Westkost [7]

Answer:

1963.2 pounds (lbs.)

Step-by-step explanation:

Things to understand before solving:

- <u>Normal Probability Distribution</u>

The z-score formula can be used to solve normal distribution problems. In a set with mean ц and standard deviation б, the z-score of a measure X is given by: $Z=\frac{X-u}{a}$

The Z-score reflects how far the measure deviates from the mean. After determining the Z-score, we examine the z-score table to determine the p-value associated with this z-score. This p-value represents the likelihood that the measure's value is less than X, or the percentile of X. Subtracting 1 from the p-value yields the likelihood that the measure's value is larger than X.

- <u>Central Limit Theorem</u>

The Central Limit Theorem establishes that, for a normally distributed random variable X, with mean ц and standard deviation б , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean ц and standard deviation $s=\frac{a}{\sqrt{n} }$

As long as n is more than 30, the Central Limit Theorem may be applied to a skewed variable. A specific kind of steel cable has an average breaking strength of 2000 pounds, with a standard variation of 100 pounds.

This means, ц = 2000 and б = 100.

A random sample of 20 cables is chosen and tested.

This means that n = 20, $s=\frac{100}{\sqrt{120} } =22.361$