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Damm [24]
3 years ago
11

Alex spends 14 days on vacation. He spends 4 out of the 14 days at the beach and half as many days visiting museums as he spends

at the beach. How many days does he spend at the beach? How many days does he spend visiting museums?
Mathematics
1 answer:
zvonat [6]3 years ago
3 0


4 days at the beach.*7 or 5* days of visiting museums.


*If you take away 4 from 14 you get 10 and halve it you get 5 but if you halve 14 you get 7*


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What is the solution to this system of equations graphed?
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Answer:

(2,-2)

Step-by-step explanation:

The solution to a system of equations like these, is where the two lines intersect. In this case, I believe, it would be at point (2,-2). Although, it is a little hard to pinpoint without the labeled number lines.

Hope this helps! Best of luck!

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The table below shows the amount of money that Bonnie spent during her last trip to Target. Target Purchases Clothing $56 Grocer
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What is the length of line segment BC?​
Nezavi [6.7K]

Answer:

d. 10.53 units

Step-by-step explanation:

This question could be answered by Pythagoras Theorem which is :

a^{2} +b^{2} =c^{2} In every triangle c is always the hypotenuse ( which is the longest line in the triangle/ the line where the right angle sign is pointing to )

a or b could be either one of the other lines of the triangle it doesn't matter.

First you find AC which is simply by adding the two 5.2 values which is 10.4 units.

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To find b you rearrange everything to the other side except b

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And if you type that in the calculator you get

BC= 10.53 units (2 decimal place)

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5 0
3 years ago
At a pet store, the lizards in aquarium A have a mean length of 13 cm and a range from 9 cm to 15 cm. In aquarium B. the
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3 0
3 years ago
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Identify the correct HYPOTHESIS used in a hypothesis test of the following claim and sample data: Claim: "The average annual hou
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Answer:

We accept H₀

Step-by-step explanation:

Population mean  μ₀ = 47500

Population standard deviation unknown

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Sample mean    μ = 48061

Sample standard deviation 2,351

The claim implies a two tail test with t-studend distributon

Null  Hypothesis                 H₀               μ  =  μ₀

Alternative Hypothesis      Hₐ               μ  ≠  μ₀      

Confidence Interval  mean α = 0,02     and   α/2  =  0,01

With α/2 and df = 85, from t-table we find  t(c)  critical value

t(c) = 2,3710

We compute t(s) as

t(s)  =  ( μ - μ₀ ) / s /√n

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t(s)  = 561 * 9,273 / 2351

t(s) = 2,212

Now we compare t(s) and t(c)

t(s) < t(c)       2,212  < 2,371

Then we are in the acceptance region. We accept H₀

5 0
3 years ago
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