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3 years ago

Answer the word problem for 10 points

Mathematics

2 answers:

sasho [114]3 years ago

6 0

Answer:

Step-by-step explanation:21

Yakvenalex [24]3 years ago

5 0

Answer:

Step-by-step explanation:

105-45=105

105/5=21

so ya

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Find the third side in simplest radical form:

larisa [96]

Answer:

x = √53

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Equality Properties

Trigonometry

[Right Triangles Only] Pythagorean Theorem: a² + b² = c²

Step-by-step explanation:

Step 1: Define

We are given a right triangle. We can use PT to solve for the missing length.

Step 2: Identify Variables

Leg a = 6

Leg b = x

Hypotenuse c = √89

Step 3: Solve for x

Set up equation: 6² + x² = (√89)²
Isolate x term: x² = (√89)² - 6²
Exponents: x² = 89 - 36
Subtract: x² = 53
Isolate x: x = √53

5 0

3 years ago

What is the total surface area?? Please help me I’m confused

Simora [160]

Answer:

216

Step-by-step explanation:

8 0

3 years ago

A company has a policy of retiring company cars; this policy looks at number of miles driven, purpose of trips, style of car and

pav-90 [236]

Answer:

ans=13.59%

Step-by-step explanation:

The 68-95-99.7 rule states that, when X is an observation from a random bell-shaped (normally distributed) value with mean $\mu$ and standard deviation $\sigma$ , we have these following probabilities

$Pr(\mu - \sigma \leq X \leq \mu + \sigma) = 0.6827$

$Pr(\mu - 2\sigma \leq X \leq \mu + 2\sigma) = 0.9545$

$Pr(\mu - 3\sigma \leq X \leq \mu + 3\sigma) = 0.9973$

In our problem, we have that:

The distribution of the number of months in service for the fleet of cars is bell-shaped and has a mean of 53 months and a standard deviation of 11 months

So $\mu = 53, \sigma = 11$

So:

$Pr(53-11 \leq X \leq 53+11) = 0.6827$

$Pr(53 - 22 \leq X \leq 53 + 22) = 0.9545$

$Pr(53 - 33 \leq X \leq 53 + 33) = 0.9973$

-----------

$Pr(42 \leq X \leq 64) = 0.6827$

$Pr(31 \leq X \leq 75) = 0.9545$

$Pr(20 \leq X \leq 86) = 0.9973$

-----

What is the approximate percentage of cars that remain in service between 64 and 75 months?

Between 64 and 75 minutes is between one and two standard deviations above the mean.

We have $Pr(31 \leq X \leq 75) = 0.9545 = 0.9545$ subtracted by $Pr(42 \leq X \leq 64) = 0.6827$ is the percentage of cars that remain in service between one and two standard deviation, both above and below the mean.