Please help Me!!!!!!!!

Help me with math please

DENIUS [597]

(-3^3+2 · -2^2)^2-4^4

(-27+2 · -4)^2-4^4

(-27+(-8))^2-4^4

-35^2-4^4

1225-256

969

8 0

3 years ago

An earthquake with a magnitude of about 2. 0 or less is called a microearthquake. It is not usually felt. The intensity of an ea

Alecsey [184]

The intensity of an earthquake with a magnitude of 2 is 100 times greater than the intensity of an a standard earthquake .

<h3>What is magnitude of earthquake?</h3>

Magnitude of earthquake is the measure of the size of origin of the earthquake. The magnitude of the earthquake keeps the same value for each place.

An earthquake with a magnitude of about 2. 0 or less is called a micro-earthquake and not felt usually.The intensity of an earthquake with a magnitude of 2.

Let the intensity of this earthquake is <em>n </em>times greater than the intensity of an a standard earthquake. Thus the intensity of standard earthquake can be given as,

$n=10^2\\n=100$

If the magnitude would be 3 then the intensity would be,

$n=10^3\\n=1000$

It would be 1000 times greater than the standard earthquake and so on.

Thus, the intensity of an earthquake with a magnitude of 2 is 100 times greater than the intensity of an a standard earthquake .

Learn more about the magnitude of earthquake here;

brainly.com/question/18109453

8 0

2 years ago

"An ordinance requiring that a smoke detector be installed in all previously constructed houses has been in effect in a particul

Galina-37 [17]

Answer:

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15) = 0.0173

b) Probability of not rejecting the claim when p = 0.7, P(X > 15) = 0.8106

when p = 0.6, P(X > 15) = 0.4246

c) Check Explanation

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Step-by-step explanation:

p is the true proportion of houses with smoke detectors and p = 0.80

The claim that 80% of houses have smoke detectors is rejected if in a sample of 25 houses, not more than 15 houses have smoke detectors.

If X is the number of homes with detectors among the 25 sampled

a) Probability that the claim is rejected when the actual value of p is 0.8 = P(X ≤ 15)

This is a binomial distribution problem

A binomial experiment is one in which the probability of success doesn't change with every run or number of trials (probability that each house has a detector is 0.80)

It usually consists of a number of runs/trials with only two possible outcomes, a success or a failure (we are sampling 25 houses with each of them either having or not having a detector)

The outcome of each trial/run of a binomial experiment is independent of one another.

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = less than or equal to 15

p = probability of success = probability that a house has smoke detectors = 0.80

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.80 = 0.20

P(X ≤ 15) = Sum of probabilities from P(X = 0) to P(X = 15) = 0.01733186954 = 0.01733

b) Probability of not rejecting the claim when p= 0.7 when p= 0.6

For us not to reject the claim, we need more than 15 houses with detectors, hence, th is probability = P(X > 15), but p = 0.7 and 0.6 respectively for this question.

n = total number of sample spaces = 25 houses sampled

x = Number of successes required = more than 15

p = probability that a house has smoke detectors = 0.70, then 0.60

q = probability of failure = probability that a house does NOT have smoke detectors = 1 - p = 1 - 0.70 = 0.30

And 1 - 0.60 = 0.40

P(X > 15) = sum of probabilities from P(X = 15) to P(X = 25)

When p = 0.70, P(X > 15) = 0.8105639765 = 0.8106

When p = 0.60, P(X > 15) = 0.42461701767 = 0.4246

c) How do the "error probabilities" of parts (a) and (b) change if the value 15 in the decision rule is replaced by 14.

The error probabilities include the probability of the claim being false.

When X = 15

(Error probability when p = 0.80) = 0.0173

when p = 0.70, error probability = P(X ≤ 15) = 1 - P(X > 15) = 1 - 0.8106 = 0.1894

when p = 0.60, error probability = 1 - 0.4246 = 0.5754

When X = 14

(Error probability when p = 0.80) = P(X ≤ 14) = 0.00555

when p = 0.70, error probability = P(X ≤ 14) = 0.0978

when p = 0.60, error probability = P(X ≤ 14) = 0.4142

The error probabilities are evidently lower when 15 is replaced with 14 in the calculations.

Hope this Helps!!!

6 0

3 years ago

What is 6 2/5 added by 2 3/10.Make sure your answer is fully reduced

slava [35]

Answer:

8 7/10

Step-by-step explanation:

6 0

2 years ago

Read 2 more answers

A pet store sells goldfish and hermit crabs,

Elis [28]

Answer:

$Cost = \$28$

Step-by-step explanation:

Given

Represent Goldfish with g and hermit crabs with h.

The first statement, we have:

$7g + 3h = 26$

The second statement, we have:

$4g + 5h = 28$

Required

Determine the selling price of 6 goldfish and 4 hermit crabs

The equations are:

$7g + 3h = 26$ --- (1)

$4g + 5h = 28$ --- (2)

Make g the subject in (2)

$4g + 5h = 28$

$4g = 28 - 5h$

Divide both sides by 4

$g = \frac{1}{4}(28 - 5h)$

Substitute $\frac{1}{4}(28 - 5h)$ for g in (1)

$7g + 3h = 26$

$7(\frac{1}{4}(28 - 5h)) + 3h = 26$

$\frac{7}{4}(28 - 5h) + 3h = 26$

Multiply through by 4

$4 * \frac{7}{4}(28 - 5h) + 4*3h = 26*4$

$7(28 - 5h) + 4*3h = 26*4$

Open bracket

$196 - 35h + 12h = 104$

$196 -23h = 104$

Collect Like Terms

$-23h = 104-196$

$-23h = -92$

Make h the subject

$h = \frac{-92}{-23}$

$h = \frac{92}{23}$

$h = 4$

Substitute 4 for h in $g = \frac{1}{4}(28 - 5h)$

$g = \frac{1}{4}(28 - 5*4)$

$g = \frac{1}{4}(28 - 20)$

$g = \frac{1}{4}(8)$

$g = 2$

This implies that:

1 goldfish = $2

1 hermit crab = $4

The cost of 6 goldfish and 4 hermit crabs is:

$Cost = 6g + 4h$

$Cost = 6*\$2 + 4*\$4$

$Cost = \$12 + \$16$

$Cost = \$28$

5 0

3 years ago