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3 years ago

(4)*(1)and move the decimal three places to the left.

Mathematics

2 answers:

enot [183]3 years ago

7 0

0.004

your welcome lol

balu736 [363]3 years ago

4 0

Answer:

0.004

Step-by-step explanation:

4 x 1

<-- move decimal 3 to the left

0.004

Brainliest?

You might be interested in

Find the rate of a discount sweater with a regular price of $39 that is on sale for $27

sammy [17]

Just do 27/39 to find how much 27 is of 39, and then subtract that from 100 to get the percentage of discount.

27/39 is about 0.69

So the rate of discount is about 0.31

Hope this helps!

6 0

3 years ago

Write 1/2 out of 10,000 as a simplified fraction and percent

12345 [234]

5000/10000
50%

i think

3 0

3 years ago

The Wilson family had 5 children. Assuming that the probability of a child being a girl is 0.5, find the probability that the Wi

Dafna11 [192]

Answer:

0.813

0.500

Step-by-step explanation:

Use binomial probability.

P = nCr p^r q^(n−r)

where n is the number of trials,

r is the number of successes,

p is the probability of success,

and q is the probability of failure (1−p).

In this problem, n = 5, p = 0.5, and q = 0.5.

"At least 2 girls" means r = 2, 3, 4, or 5.

Or, we can use the complement.

P(at least 2 girls) = 1 − P(at most 1 girl)

P(at least 2 girls) = 1 − P(r=0 or r=1)

P(at least 2 girls) = 1 − ₅C₁ (0.5)¹ (0.5)⁵⁻¹ − ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰

P(at least 2 girls) = 1 − 5 (0.5) (0.5)⁴ − 1 (1) (0.5)⁵

P(at least 2 girls) = 1 − 6 (0.5)⁵

P(at least 2 girls) ≈ 0.813

"At most 2 girls" means r = 0, 1, or 2.

P(at most 2 girls) = P(r=0, r=1, or r=2)

P(at most 2 girls) = ₅C₀ (0.5)⁰ (0.5)⁵⁻⁰ + ₅C₁ (0.5)¹ (0.5)⁵⁻¹ + ₅C₂ (0.5)² (0.5)⁵⁻²

P(at most 2 girls) = 1 (1) (0.5)⁵ + 5 (0.5) (0.5)⁴ + 10 (0.5)² (0.5)³

P(at most 2 girls) = 16 (0.5)⁵

P(at most 2 girls) = 0.500

4 0

3 years ago

Brian invests £4900 into his bank account.

True [87]

Answer:

£5200.68

Step-by-step explanation:

<u>Given:</u>

Investment P = £4900
Interest rate r = 1.5% or r = 0.015
Time t = 4 years
Number of compounds per year n = 1

<u>Find the future amount:</u>

$F = P*(1+r/n)^{nt}\\$
$F = 4900*(1+0.015/1)^{1*4}=4900*1.015^4 = 5200.68$

6 0

2 years ago

Read 2 more answers

suppose X and Y are independent random variables, both with normal distributions. If X has a mean of 45 with a standard deviatio

djyliett [7]

Answer:

0.9772 = 97.72% probability that a randomly generated value of X is greater than a randomly generated value of Y

Step-by-step explanation:

When the distribution is normal, we use the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu_X = 45, \sigma_X = 4, \mu_Y = 35, \sigma_Y = 3$

What is the probability that a randomly generated value of X is greater than a randomly generated value of Y

This means that the subtraction of X by Y has to be positive.

When we subtract two normal variables, the mean is the subtraction of their means, and the standard deviation is the square root of the sum of their variances. So

$\mu = \mu_X - \mu_Y = 45 - 35 = 0$