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german
3 years ago
7

QUICK QUICK PLEASE! 1 MINUTE

Mathematics
2 answers:
Dima020 [189]3 years ago
4 0

Step-by-step explanation:

Total tomato plants = 24 + 30 = 54

Total green bean plants = 18 + 25 = 43

The ratio of tomato plants to green bean plants

=  \frac{54}{43}  = 54 \:  :  \: 43

kirill115 [55]3 years ago
4 0

Answer:

The answer is different, sorry for the late response

Step-by-step explanation:

You might be interested in
calvin and susie baked some cookies in the ratio of 3 : 7. calvin baked 60 fewer cookies than susie. how many cookies did they b
kow [346]

Answer:

Step-by-step explanation:

Ratio of cakes baked by Calvin and Susie = 3 : 7

Cakes baked by Calvin = 3x

Cakes baked by Susie = 7x

Cakes baked by Calvin = Cakes baked by Susie - 60

3x = 7x - 60

7x - 60 = 3x

7x        = 3x  + 60

7x - 3x = 60

      4x  = 60

        x = 60/4

x = 15

Cakes baked by Calvin = 3*15 = 45

Cakes baked by Susie = 7 *15 = 105

Total cakes baked = 45 + 105 = 150

They both baked 150 cakes altogether.

8 0
2 years ago
Can someone explain how to graph this? I know how to graph it in the y= format. So can someone explain how to get these equation
artcher [175]

Answer:

hope this helps.

Step-by-step explanation:

You don't really need to get them in the y format.

make a table and plug some numbers in and graph those points

use a ruler to connect the dots.

y = -3 is a horizontal line

8 0
3 years ago
The process standard deviation is 0.27, and the process control is set at plus or minus one standard deviation. Units with weigh
mr_godi [17]

Answer:

a) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

b) P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

c) For this case the advantage is that we have less items that will be classified as defective

Step-by-step explanation:

Assuming this complete question: "Motorola used the normal distribution to determine the probability of defects and the number  of defects expected in a production process. Assume a production process produces  items with a mean weight of 10 ounces. Calculate the probability of a defect and the expected  number of defects for a 1000-unit production run in the following situation.

Part a

The process standard deviation is .15, and the process control is set at plus or minus  one standard deviation. Units with weights less than 9.85 or greater than 10.15 ounces  will be classified as defects."

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(10,0.15)  

Where \mu=10 and \sigma=0.15

We can calculate the probability of being defective like this:

P(X

And we can use the z score formula given by:

z=\frac{x-\mu}{\sigma}

And if we replace we got:

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.15}) = P(Z>1)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.159+0.159 = 0.318

And the expected number of defective in a sample of 1000 units are:

X= 0.318*1000= 318

Part b

Through process design improvements, the process standard deviation can be reduced to .05. Assume the process control remains the same, with weights less than 9.85 or  greater than 10.15 ounces being classified as defects.

P(X

And for the other case:

tex] P(X>10.15)[/tex]

P(X>10.15)= P(Z > \frac{10.15-10}{0.05}) = P(Z>3)=1-P(Z

So then the probability of being defective P(D) is given by:

P(D) = 0.00135+0.00135 = 0.0027

And the expected number of defective in a sample of 1000 units are:

X= 0.0027*1000= 2.7

Part c What is the advantage of reducing process variation, thereby causing process control  limits to be at a greater number of standard deviations from the mean?

For this case the advantage is that we have less items that will be classified as defective

5 0
3 years ago
Find the slope and the x and y-intercept of the equation 2x-3y=12
Maslowich
X=9
Y=2 i think it’s the answer
4 0
3 years ago
john bought a full basket of mangoes when he got home he realised 12 out of the 20 mangoes he bought are bad what is the probabi
lara31 [8.8K]
12/20 mangoes are bad so 8/20 mangoes are good and to convert to a percentage we make the denominator 100.
To make 20 to 100 we times by 5 so we do the same to the numerator which is 8
So 8x5 = 40%
Give me brainliest answer plz
And hope it helps!
6 0
2 years ago
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