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rewona [7]
3 years ago
10

Julia wants to bring watermelon to the party. One serving of watermelon weigh about 150 grams. There will be 60 people at the pi

cnic. How many kilograms of watermelon will Julia need to buy?
Mathematics
2 answers:
Ulleksa [173]3 years ago
5 0

we are given

One serving of watermelon weigh about 150 grams

watermelon required by one person =150 g

total number of persons =60

total amount of watermelon = total number of persons*watermelon required by one person

total amount of watermelon=60*150g

total amount of watermelon=9000g

we know that

1000g=1kg

we can write as

total amount of watermelon=9*1000g

we can replace 1000g as 1kg

total amount of watermelon=9 kg...........Answer

lilavasa [31]3 years ago
3 0
She will need to buy 9 kg of watermelon.

150g(16) = 9000 g that she needs; each kilogram is 1000 grams:
9000/1000 = 9
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Whats the common factor of 18,27
Leviafan [203]

Well, it depends, there may be multiple common factors, if it's the greatest common factor, then there is only one.  The way to do this is to list all the factors of each number.


So Factors of

18: 1,2,3,6,9,18

27: 1,3,9,27

Common Factors : 1, 3, 9

Greatest Common Factor: 9

5 0
3 years ago
It is thought that not as many Americans buy presents to celebrate Valentine's Day anymore. A random sample of 40 Americans yiel
evablogger [386]

Answer:

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

 (0.3672 , 0.7328)

Step-by-step explanation:

<u><em>Explanation:</em></u>-

Given Random sample size n =40

Sample proportion

                            p = \frac{x}{n} = \frac{22}{40} = 0.55

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

   

          (p-Z_{\alpha } \sqrt{\frac{pq}{n} } , p + Z_{\alpha } \sqrt{\frac{pq}{n} } )

         

The Z-value Z₀.₉₈ = Z₀.₀₂ = 2.326

98% confidence interval of the true proportion of all Americans who celebrate Valentine's Day

    (0.55-2.326\sqrt{\frac{0.55 X0.45}{40} } , 0.55 + 2.326\sqrt{\frac{0.55 X0.45}{40} } )

  ( 0.55 - 0.1828 , 0.55 + 0.1828)

  (0.3672 , 0.7328)

         

6 0
3 years ago
Read 2 more answers
Help me with this plz, I ain’t good at math at all.
Ad libitum [116K]
I’ll be happy to help!
Since a right angle is equal to 90 degrees, we can add the 67+90 which equals 157, so since a triangle is equal to 180, we can subtract 180-157.
180-157= 23
The answer for x is 23!
5 0
3 years ago
PLEASE HELP ME!
ELEN [110]

Answer:

the answer is c

Step-by-step explanation:

8 0
3 years ago
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Can somebody please help m eits 9th grade math im so confused !
prisoha [69]

Answer:

Option 3: (5, 1)

Step-by-step explanation:

There are several ways of solving for systems of linear equations, either through graphing, substitution, or elimination.  

The graphing method is the easiest, as you only need to plot points on the graph using the y-intercepts and the slopes.  

I will be demonstrating the <u>graphing method</u>.  

Given the systems of linear equations:  

Equation 1:  y = x - 4

where the <u>slope</u>, <em>m</em> = 1, and <u>y-intercept </u>(0, -4).

Equation2: y = -x + 6

where the <u>slope</u>, <em>m </em>= -1, and <u>y-intercept</u> (0, 6).

The y-intercept is the point on the graph where it crosses the y-axis, and has coordinates of (0, b).

<h3>Graphing Instructions:</h3>

For Equation 1, plot the y-intercept, (0, -4) on the graph. Then use the <u>slope</u>, m = 1 (rise 1 unit, run 1 unit) to plot other points on the graph. Repeat this process until you have enough points to connect a line with.  

Repeat these same procedures for graphing Equation 2, start by plotting its y-intercept at (0, 6), and its slope, m = -1 (down 1 unit, run 1 unit).

<h3>Solution: </h3>

The intersection of both lines occurs at point, (5, 1), which is solution to the given systems of linear equations.  Therefore, the correct answer is the third option, (5, 1).

Attached is the screenshot of the graphed systems of linear equations, where it shows the point of intersection at (5, 1).  

5 0
3 years ago
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