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LenKa [72]
3 years ago
6

Write the value of the ratio of the amount of total sugar to the amount of white sugar

Mathematics
1 answer:
lyudmila [28]3 years ago
3 0
Right after you finish writing this question.
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He sum of three numbers is 19 . the sum of twice the first​ number, 4 times the second​ number, and 5 times the third number is
Alja [10]
Let 
x----------> the first number
y---------> the second number
z--------> the third number

we know that
x+y+z=19-----> z=19-x-y------> equation 1
2x+4y+5z=72-----> equation 2
4x-y=12------> equation 3

substitute equation 1 in 2
2x+4y+5*[19-x-y]=72----> 2x+4y+95-5x-5y=72----> -3x-y=-23
3x+y=23------> equation 4

add equation 3 and equation 4
4x-y=12
3x+y=23
--------------
7x+0y=35--------> x=35/7----------> x=5
4x-y=12------> y=4x-12----> y=4*5-12----> y=8
z=19-x-y------> z=19-5-8----> z=6

the answer is
the numbers are
5 
8
6


6 0
4 years ago
81x^2 = 4 NEED HELP SOLVING QUADRATIC EQAUTIONS
ser-zykov [4K]
X=-+.222222 this is what I calculated
8 0
4 years ago
Read 2 more answers
The number of years since Lena graduated
sweet-ann [11.9K]
No it’s not proportional. If you had a coefficient in front of the a it could have been proportional. But the addition or subtraction of a quantity leads it to be nonproportional
3 0
3 years ago
According to a study done by Wakefield Research, the proportion of Americans who can order a meal in a foreign language is 0.47.
UNO [17]

Answer:

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

Step-by-step explanation:

We are given that according to a study done by Wake field Research, the proportion of Americans who can order a meal in a foreign language is 0.47.

Suppose a random sample of 200 Americans is asked to disclose whether they can order a meal in a foreign language.

<em>Let </em>\hat p<em> = sample proportion of Americans who can order a meal in a foreign language</em>

The z-score probability distribution for sample proportion is given by;

          Z = \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion

p = population proportion of Americans who can order a meal in a foreign language = 0.47

n = sample of Americans = 200

Probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is given by = P( \hat p > 0.50)

  P( \hat p > 0.06) = P( \frac{ \hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } > \frac{0.5-0.47}{\sqrt{\frac{0.5(1-0.5)}{200} } } ) = P(Z > 0.85) = 1 - P(Z \leq 0.85)

                                                               = 1 - 0.80234 = <u>0.19766</u>

<em>Now, in the z table the P(Z  </em>\leq <em>x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 0.85 in the z table which has an area of 0.80234.</em>

Therefore, probability that the proportion of Americans who can order a meal in a foreign language is greater than 0.5 is 0.19766.

4 0
4 years ago
Three football teams scored a total of 256 points for the season. The Robins scored 35 more points than the Jays. Complete the t
Phoenix [80]

Answer:

Robins= j+35

(J-63)+j+(j+35)=256

Step-by-step explanation:

3 0
4 years ago
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