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tatuchka [14]
3 years ago
11

If n orange juice is made up of concentrate and water in the ratio 3:8. What fraction of the mixture is concentrate?​

Mathematics
1 answer:
enot [183]3 years ago
5 0

Answer:your answer should be 2

Step-by-step explanation:

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Bob has some 10 lb weights and some 3 lb weights. Together, all his weights add up to 50 lbs. If x represents the number of 3 lb
Romashka [77]
Total weight = 50 lb
x = number of 3-lb weights
y = number of 10-lb weights

weight of 3-lb weights = 3x
weight of 10-lb weights = 10y
total weight = 3x + 10y

equation
3x + 10y = 50
8 0
3 years ago
Find the number that makes the ratio equivalent to 7:4.
MariettaO [177]

Answer:

Any number: 49:28

Step-by-step explanation:

Whatever number you multiply, it will still simplify to 7:4

4 0
3 years ago
Evaluate 3x-2y; if x=7 and y=3
Zarrin [17]

Answer:

15

Step-by-step explanation:

3x - 2y = ?

3(7) - 2(3) = ?

21 - 6 = ?

= 15

All you have to do is substitute the given values & follow the order of operations! Best of luck.

3 0
3 years ago
5x-2=3x+40 answer plz :')
noname [10]

X=21

5x(-3x) -2= 3x(-3x) +40

2x-2(+2)=40(+2)

2x=42

X=21

7 0
3 years ago
Philips Semiconductors is a leading European manufacturer of integrated circuits. Integrated circuits are produced on silicon wa
shepuryov [24]

Answer:

Step-by-step explanation:

1) The null hypothesis is,

H_0: The mean thickness of teh wafers for the five positions are equal

i.e, H_0:\mu_1=\mu_2=\mu_3=\mu_4=\mu_5

2)

The alternative hypothesis is,

H_1: There is an evidence of a difference in the mean thickness of the wafers for the five positions

3)

Let us consider the level of significance \alpha=0.01

from the Minitab outout

One-way ANOVA:C1 versus C2

source         DF            SS                 MS               F            P

C2                  4      1417.73          354.43        51.00       0.00

Error           145     1007.77              6.95

Total           14      2425.50

S = 2.636       R - S = 58.45%     R - Sq(adj) = 57.31%

Individual 95% CIs For Mean Based on Pooled StDev

level         N         Mean            StDev    -,----------,----------,----------,----------

1              30    240.53                2.62    (--,--)

2             30     243.73                2.79             (--,--)

3             30     246.07                2.90                           (--,--)

4             30     249.10                 2.66                                        (--,--)

5             30     247.07                 2.15                                  (--,--)

                                                               -,----------,----------,----------,----------

                                                        240.0   243.0   246.0   249.0

Pooled StDev = 2.64

4)

The test statistic is, F = 51

5)

The P-value is approximately 0

6)

Here, the P - value is less than the level of significance

\therefore \,P-value

So, we do not accept our null hypothesis H_0

7)

Therefore, we conclude that there is an evidence of a difference in the mean thickness of the wafers for the five positions at level of significance \alpha=0.05

b)

chek attachment

we observe that,

The mean thickness of the wafer for position 1 is significant with position 2,

position 18, position 19 and position 28.

The mean thickness of the wafer for position 2 is significant with position 18,

position 19 and position 28.

The mean thickness of the wafer for position 18 is significant with the

position 19.

But the mean thickness of the wafer for position18 is not significant with

position 28.

The mean thickness of the wafer for position 19 is significant with position 28.

7 0
3 years ago
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