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FinnZ [79.3K]
3 years ago
5

Danielle can type at a rate of 55 words per minute. How many words can she type in four and a half minutes?​

Mathematics
1 answer:
Romashka [77]3 years ago
6 0
55x4=220+55➗2=27.5

Therefore the answer is 247.5
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Of equations to answer the questions.
iren [92.7K]
2x + 3(8 - 3x) = 3
2x + 24 -9x = 3
-7x = -21
x = 3

y = 8 - 3x
y = 8 - 3(3)
y = 8-9
y = -1

Check the math:
2(3) + 3(-1) = 6 - 3 = 3
4 0
3 years ago
Elevii clasei a IV-a au fost la patinoar. Stiind ca numarul baietilor reprezinta jumatate plus 3 din numarul copiilor si ca numa
Shalnov [3]

Step-by-step explanation:

Fie x numărul de copii

f = 10 =\:fete

b\:=\:\frac{x}{2}\:+\:3

f\:+\:b\:=\:x

10\:+\:\frac{x}{2}\:+\:3\:=\:x

20\:+\:x\:+\:6\:=\:2x

x\:=\:26\:copii\:au\:fost\:la\:patinoar

b=\frac{26}{2}+\:3

b\:=\:16\:baieti

7 0
3 years ago
The price of a sandwich is $1.50 more than the price of a smoothie, which is d dollars. What does the expression d+1.5 represent
liberstina [14]

Answer:

The expression d+1.5 represents the price of the sandwich, d in the scenario represents the smoothie price.

7 0
3 years ago
Read 2 more answers
Select the equation where x = 6 is a solution.
olga55 [171]

Answer: D 5x=30

Step-by-step explanation

You just substitute 6 into the equations to see if they make the equation true.

8 0
3 years ago
Read 2 more answers
The distribution of weekly salaries at a large company is right skewed with a mean of $1000 and a standard deviation of $350. Wh
Shtirlitz [24]

Answer:

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

Step-by-step explanation:

To solve this question, we have 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 random variable X, with mean \mu and standard deviation \sigma, a large sample size can be approximated to a normal distribution with mean \mu and standard deviation, which is also called standard error s = \frac{\sigma}{\sqrt{n}}

In this problem, we have that:

\mu = 1000, \sigma = 350, n = 50, s = \frac{350}{\sqrt{50}} = 49.5

What is the probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

This is the pvalue of Z when X = 1000+50 = 1050 subtracted by the pvalue of Z when X = 1000-50 = 950. So

X = 1050

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{1050-1000}{49.5}

Z = 1.01

Z = 1.01 has a pvalue of 0.8438

X = 950

Z = \frac{X - \mu}{s}

Z = \frac{950-1000}{49.5}

Z = -1.01

Z = -1.01 has a pvalue of 0.1562

0.8438 - 0.1562 = 0.6876

68.76% probability that the sampling error made in estimating the mean weekly salary for all employees of the company by the mean of a random sample of weekly salaries of 50 employees will be at most $50

4 0
3 years ago
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