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grandymaker [24]
2 years ago
6

Translate this phrase into an algebraic expression. 73 less than twice Jose's height Use the variable to represent Jose's height

.​
Mathematics
1 answer:
Luba_88 [7]2 years ago
8 0

Answer:

x-73

Step-by-step explanation:

Let x=Jose's height.  If it says "less than," then that is subtraction.  Since Jose's height is not defined, there is no specific number that can be used to describe "73 less than twice Jose's height," so I use the variable x, and then subtract 73.

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147.1 is what percent of 155
Harrizon [31]
Solution for what is 147.1% of 155

155/x=100/147.1
(155/x)*x=(100/147.1)*x       - we multiply both sides of the equation by x
155=0.67980965329708*x       - we divide both sides of the equation by (0.67980965329708) to get x
155/0.67980965329708=x 
228.005=x 
x=228.005

now we have: 
147.1% of 155=228.005

3 0
2 years ago
The radioactive element carbon-14 has half of 5750 years. A scientist determined that the bones from a mastodon has lost 56.6% o
IRISSAK [1]

Answer:7881 years

Step-by-step explanation:

You can use the equation

fraction remaining = 0.5n where n is # of half lives elapsed

n = ?

fraction remaining = 1 - 0.612 = 0.388

0.388 = 0.5n

n = 1.37 = the # of half lives

1.37 x 5750 yrs = 7881 years

8 0
3 years ago
I need help figuring out this question
DiKsa [7]

Answer:

no picture

Step-by-step explanation:

4 0
2 years ago
Researchers at the National Cancer Institute released the results of a study that investigated the effect of weed-killing herbic
jolli1 [7]

Answer:

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

Step-by-step explanation:

Before building the confidence interval, we need to understand the central limit theorem and subtraction between normal variables.

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 a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

The randomly sampled 400 dogs from homes where an herbicide was used on a regular basis, diagnosing lymphoma in 230 of them.

This means that:

p_h = \frac{230}{400} = 0.575, s_h = \sqrt{\frac{0.575*0.425}{400}} = 0.0247

Of 200 dogs randomly sampled from homes where no herbicides were used, only 25 were found to have lymphoma.

This means that:

p_n = \frac{25}{200} = 0.125, s_n = \sqrt{\frac{0.125*0.875}{200}} = 0.0234

Distribution of the difference:

p = p_h - p_n = 0.575 - 0.125 = 0.45

s = \sqrt{s_h^2+s_n^2} = \sqrt{0.0247^2 + 0.0234^2} = 0.034

Confidence interval:

The confidence interval is:

p \pm zs

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

95% confidence level

So \alpha = 0.05, z is the value of Z that has a pvalue of 1 - \frac{0.05}{2} = 0.975, so Z = 1.96.

The lower bound is 0.45 - 1.96(0.034) = 0.3834

The upper bound is 0.45 + 1.96(0.034) = 0.5166

The 95% confidence interval for the difference in the proportion of cancer diagnoses between the two groups is (0.3834, 0.5166).

3 0
3 years ago
Charlie ran 5 miles in 51.5 minutes. Use the equation 5r=51.5 to find r, the average rate he ran.
jenyasd209 [6]
5r = 51.5
r = 51.5 / 5
r = 10.3
6 0
2 years ago
Read 2 more answers
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