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spayn [35]
2 years ago
8

How many oranges does Heidi need to make one large orange juice? Show your work.

Mathematics
1 answer:
Sindrei [870]2 years ago
4 0

Answer:

Heidi needs to make 9 oranges to make one large orange juice, give me brainliest answer:)

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a store owner wishes to make a new tea with a unique flavor by mixing black tea and oolong tea. if he has 35 pounds of oolong te
Verizon [17]

35 pounds of black tea is needed to be mixed to get the final mixture for 2.10 per pound

<h3>What is an equation?</h3>

An equation is an expression that shows the relationship between two or more numbers and variables.

Let x represent the amount of black tea worth 1.80 per pound to be mixed to get the final mixture for 2.10 per pound, hence:

2.4(35) + 1.8(x) = 2.1(x + 35)

x = 35 pounds

35 pounds of black tea is needed to be mixed to get the final mixture for 2.10 per pound

Find out more on equation at: brainly.com/question/2972832

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3 0
1 year ago
Help me please!!! help!
Alenkinab [10]

Answer:

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Step-by-step explanation:


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2 years ago
Simplify 3/4 - 1 1/4
lbvjy [14]
I think the answer is -7/4.
4 0
3 years ago
Read 2 more answers
One stats class consists of 52 women and 28 men. Assume the average exam score on Exam 1 was 74 (σ = 10.43; assume the whole cla
Svetllana [295]

Answer:

(A) What is the z- score of the sample mean?

The z- score of the sample mean is 0.0959

(B) Is this sample significantly different from the population?

No; at 0.05 alpha level (95% confidence) and (n-1 =79) degrees of freedom, the sample mean is NOT significantly different from the population mean.

Step -by- step explanation:

(A) To find the z- score of the sample mean,

X = 75 which is the raw score

¶ = 74 which is the population mean

S. D. = 10.43 which is the population standard deviation of/from the mean

Z = [X-¶] ÷ S. D.

Z = [75-74] ÷ 10.43 = 0.0959

Hence, the sample raw score of 75 is only 0.0959 standard deviations from the population mean. [This is close to the population mean value].

(B) To test for whether this sample is significantly different from the population, use the One Sample T- test. This parametric test compares the sample mean to the given population mean.

The estimated standard error of the mean is s/√n

S. E. = 16/√80 = 16/8.94 = 1.789

The Absolute (Calculated) t value is now: [75-74] ÷ 1.789 = 1 ÷ 1.789 = 0.559

Setting up the hypotheses,

Null hypothesis: Sample is not significantly different from population

Alternative hypothesis: Sample is significantly different from population

Having gotten T- cal, T- tab is found thus:

The Critical (Table) t value is found using

- a specific alpha or confidence level

- (n - 1) degrees of freedom; where n is the total number of observations or items in the population

- the standard t- distribution table

Alpha level = 0.05

1 - (0.05 ÷ 2) = 0.975

Checking the column of 0.975 on the t table and tracing it down to the row with 79 degrees of freedom;

The critical t value is 1.990

Since T- cal < T- tab (0.559 < 1.990), refute the alternative hypothesis and accept the null hypothesis.

Hence, with 95% confidence, it is derived that the sample is not significantly different from the population.

6 0
3 years ago
Which option below best describes the polynomial 10x^3 ?
MAXImum [283]
The answer is C, there’s only 1 term= monomial
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