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2 years ago

15 crackers weigh 69 grams. how many kilograms is this?

Mathematics

1 answer:

Marrrta [24]2 years ago

6 0

Answer:

$0.069\ kg$

Step-by-step explanation:

we know that

1 kg=1,000 g

using proportion

Find out how many kilograms are 69 grams

Let

x -----> the weight in kg

$\frac{1}{1,000}\frac{kg}{g} =\frac{x}{69}\frac{kg}{g}\\\\x=\frac{69}{1,000}\\\\x=0.069\ kg$

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Prime factors for 144

matrenka [14]

Answer:

2×2×2×2×3×3 = 144

hope that helps uhh!!

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1 year ago

Read 2 more answers

9^2= 9.5^2 + 4^2 - 2 (9.5) (4) (cosR) I'm trying to find out the angle measure of R

alexandr402 [8]

Sorry I can't solve it.

Firstly, the equation is wrong. It is supposed to be 81 ≠ 30.25

Second of all, this is the best I got 81= −76 (cos(r) + 106.25

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2 years ago

The SAT is standardized to be normally distributed with a mean µ = 500 and a standard deviation σ = 100. What percentage of SAT

Aliun [14]

Answer:

34.134%

68.268%

Step-by-step explanation:

Given that:

Mean (m) = 500

Standard deviation (s) = 100

Percentage between 500 and 600

P(500 < x < 600)

P(x < 600) - P(x < 500)

Z = (x - m) / s

P(x < 600)

Z = (600 - 500) /100 = 1

P(x < 500)

Z = (500 - 500) / 500 = 0

P(Z< 1) - P(Z < 0)

0.84134 - 0.5

= 0.34134

= 0.34134 * 100%

= 34.134%

B.) Between 400 and 600

P(x < 400)

Z = (400 - 500) /100 = - 1

P(x < 600)

Z = (600 - 500) / 500 = 1

P(Z< 1 ) - P(Z < - 1)

0.84134 - 0.15866

= 0.68268

= 0.68268 * 100%

= 68.268%

8 0

2 years ago

Suppose that 20% of the residents in a certain state support an increase in the property tax. An opinion poll will randomly samp

aleksklad [387]

Answer:

95.44% probability the resulting sample proportion is within .04 of the true proportion.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using 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.

Central Limit Theorem

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

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For the sampling distribution of the sample proportion in sample of size n, the mean is $\mu = p$ and the standard deviation is $s = \sqrt{\frac{p(1-p)}{n}}$