Choose the answer. 3. (1 pt) Which is an example of the distributive property? A. 9 • (t + 5) = 9t + 45 B. 9 • (t + 5) = (t + 5)
2 answers:
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32.3 is a decimal number in which 3 is the digit at the tenth place.
Mixed number contains a whole number and a fraction part.
32.3 expressed as mixed fraction is , here 32 is the whole number and
is the fraction part.
is the fraction in the simplest form.
For this reason, 32.3 expressed as a mixed number in the simplest form is .
We know that the percentage amount of any number can be found by first converting the the given percentage to a fraction and then multiplying it by the given number. Thus, in our case, the exact value, that is 48% of 1165 can be calculated as follows:
We know that
Therefore, 48% of 1165 can be re-written as:
This can be calculated as [tex]0.48\times 1165=559.2[/tex
Therefore The exact value that is 48% of 1165 is 559.2
Answer:
Step-by-step explanation:
Hello!
The research team created a cracker with fewer calories. The average content of calories of the new crackers per serving of 6 should be less than 60.
To test it a random sample of 26 samples of the new cracker was taken and the calories per serving were measured.
Then the study variable is
X: calories of a 6 serve sample of the new reduced-calorie version. (cal)
The variable has a normal distribution with a population standard deviation of 0.82 cal.
To test the claim that the new crackers have on average less than 60 calories, the parameter of interest is the population mean (μ) and the hypotheses are:
H₀: μ ≥ 60
H₁: μ < 60
α: 0.01
Since the variable has a normal distribution and the population variance is known, the best statistic to use to conduct the test is a Standard Normal
This test is one tailed to the left, wich means that the null hypothesis will be rejected at low levels of the statistic.
If Z ≤ -2.334, the decision is to reject the null hypothesis.
If Z > -2.334, the decision is to not reject the null hypothesis.
Using the data of the sample I've calculated the sample mean.
X[bar]= ∑X/n= 1548.61/26= 59.56 cal
The observed Z value is less than the critical value, so the decision is to reject the null hypothesis.
At a level of significance of 1%, you can conclude that the population mean of calories of the samples of new crackers is less than 60 cal.
I hope it helps!
Answer:
option 3
Step-by-step explanation: