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3 years ago

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Lilianna has $68. She needs at least $194 for a new cell phone. Write an inequality to fit this situation. What's the least amou

nt of additional money she needs to buy the new cell phone?

1 answer:

Ne4ueva [31]3 years ago

8 0

Answer:

The least amount of additional money that Lilianna needs is $\$126$

Step-by-step explanation:

Let

x------> the least amount of additional money

we know that

$x+\$68\geq\$194$ -----> inequality that represent the situation

Solve for x

$x\geq\$194-\$68$

$x\geq\$126$

The minimum amount of money that Lilianna needs is $$126$

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According to a survey, 65% of murders committed last year were cleared by arrest or exceptional means. Fifty murders committed

monitta

Answer:

a) $P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421$

b) $P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714$

$P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502$

$P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319$

And adding these values we got:

$P(36 \leq X \leq 38)= 0.1535$

c) We can find the expected value given by:

$E(X) = np =50*0.65 = 32.5$

And the standard deviation would be:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

$\mu -2\sigma = 32.5- 2*3.373 = 25.75$

And then we can consider a value of 18 as unusual lower for this case.

Step-by-step explanation:

Let X the random variable of interest "number cleared by arrest or exceptional", on this case we can model this variable with this distribution:

$X \sim Binom(n=50, p=0.65)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

Part a

We want this probability:

$P(X=41)=(50C41)(0.65)^{41} (1-0.65)^{50-41}=0.00421$

Part b

We want this probability:

$P(36 \leq X \leq 38)$

We can find the individual probabilities:

$P(X=36)=(50C36)(0.65)^{36} (1-0.65)^{50-36}=0.0714$

$P(X=37)=(50C37)(0.65)^{37} (1-0.65)^{50-37}=0.0502$

$P(X=38)=(50C38)(0.65)^{38} (1-0.65)^{50-38}=0.0319$

And adding these values we got:

$P(36 \leq X \leq 38)= 0.1535$

Part c

We can find the expected value given by:

$E(X) = np =50*0.65 = 32.5$

And the standard deviation would be:

$\sigma = \sqrt{np(1-p)} \sqrt{50*0.65*(1-0.65)}= 3.373$

We can use the approximation to the normal distribution and we have at leat 95% of the data within 2 deviations from the mean. And the lower limit for this case would be:

$\mu -2\sigma = 32.5- 2*3.373 = 25.75$

And then we can consider a value of 18 as unusual lower for this case.

6 0

3 years ago

Round 89,659 to the nearest hundred thousand

Nesterboy [21]

89,659 is rounded 100,000 hundred thousand

8 0

2 years ago

Expand the following:<br> a) x(x - 6)<br> b) x(4x + 1)<br> c) 2x(5x + 4)<br> d) 3x(5x - y)

puteri [66]

Answer:

<h2>[a]=x²_6x</h2><h2>[b]=4x²+x</h2><h2>[c]=10x²+8x</h2><h2>[d]15x²_3xy</h2>

Step-by-step explanation:

5 0

3 years ago

Read 2 more answers

A rectangle is 6m wide and 11m long. How long is the diagonal of the rectangle?

Svet_ta [14]

Answer:

≈12.53

Step-by-step explanation:

7 0

3 years ago

Suppose a batch of metal shafts produced in a manufacturing company have a population standard deviation of 1.3 and a mean diame

lbvjy [14]

Answer:

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

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.

In this problem, we have that:

$\mu = 208, \sigma = 1.3, n = 60, s = \frac{1.3}{\sqrt{60}} = 0.1678$

What is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

Lesser than 208 - 0.1 = 207.9 or greater than 208 + 0.1 = 208.1. Since the normal distribution is symmetric, these probabilities are equal, so we find one of them and multiply by 2.

Lesser than 207.9.

pvalue of Z when X = 207.9. So

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

By the Central Limit Theorem

$Z = \frac{207.9 - 208}{0.1678}$

$Z = -0.6$

$Z = -0.6$ has a pvalue of 0.2743

2*0.2743 = 0.5486

54.86% probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.1 inches

6 0

2 years ago