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Veseljchak [2.6K]
3 years ago
15

What is the GCF of 42 and 91?

Mathematics
1 answer:
Neko [114]3 years ago
7 0

Answer:

7

Step-by-step explanation:

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The answer is 480. Because you times 200 by 6, you also times the number of male elephants with tusks (80) by 6. That means the answer is 480.
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Help 1-4 please 94 points
Mice21 [21]
Q1=2 because .5+.5+.5+.5=2
Q2=3 because 3/4 times 4=3
Q3=6 because all of the 1/4 added together =1 plus Q1 and Q2=6
Q4=6 divided by the ammout of bags.
7 0
4 years ago
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Given that the discriminant of a quadratic equation is 0, determine the number of real solutions.
Alisiya [41]
If the discrimination is equal to 0, there is one real solution
8 0
3 years ago
Ed buys a box of eggs costing £2.20, two packs of bacon for £2.80 each and two tins of baked beans.
Vika [28.1K]

Solving a linear equation, we can conclude that each tin costs £0.75

<h3>How much does a tin of beans cost in pounds?</h3>

First, we know that Ed buys one box of eggs for £2.20.

Also two packs of bacon for £2.80 each.

And two tins of beans for X each.

Then the total cost is:

C =  £2.20 + 2*£2.80 + 2*X

Now, we also know that when he pays with £10, he gets a change of £0.70.

Then we can solve the linear equation:

£10 - (£2.20 + 2*£2.80 + 2*X) = £0.70

£2.20 - 2*X = £0.70

£2.20 - £0.70 = 2*X

£1.50 = 2*X

£1.50/2 = X = £0.75

In this way, we can conclude that each tin costs £0.75

If you want to learn more about linear equations:

brainly.com/question/1884491

#SPJ1

4 0
2 years ago
A local newspaper was wondering if its average reader lived more than 25 miles from its headquarters. The newspaper surveyed its
Alex777 [14]

Answer:

We fail to reject H₀.

At the 5% significance level, the data do not provide sufficient evidence to conclude that the average distance a reader lives from the newspaper's headquarters is greater than 25 miles.

Step-by-step explanation:

The hypothesis for the statistical test is defined as follows:

<em>H</em>₀ : <em>μ </em>=25 vs. <em>Hₐ</em> : <em>μ </em>> 25

The test statistic value is, <em>z</em>₀ = 0.22.

The <em>p</em>-value of the test is, <em>p</em>-value = 0.41.

The significance level is, <em>α</em> = 0.05.

The <em>p</em>-value is well defined as per the probability, [under the null hypothesis (H₀)], of attaining a result equivalent to or more extreme than what was truly observed.

A small <em>p</em>-value (typically ≤ 0.05) specifies strong evidence against the null hypothesis (H₀), so you discard H₀.

A large p-value (> 0.05) specifies fragile proof against the H₀, so you fail to discard H₀.

Here, <em>p</em>-value = 0.41 > <em>α</em> = 0.05.

The null hypothesis was failed to be rejected at 5% level of significance.

Conclusion:

There is not enough evidence to support the claim that the average distance a reader lives from the newspaper's headquarters is greater than 25 miles.

6 0
3 years ago
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