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erastova [34]
3 years ago
7

6 4/11 x2 1/12 Estimate the product

Mathematics
1 answer:
Aleksandr-060686 [28]3 years ago
7 0

Answer:

The estimated product is 12.

Step-by-step explanation:

When rounding fractions, we can round them to 0, 1/2 or 1. 6 4/11 can be rounded to 6 since 4/11 isn't near the half of 11. 2 1/12 can be rounded to 2 since 1/12 is obviously close to 0. 6 x 2 = 12. The estimated product is 12.

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I just need help to solve these steps.
My name is Ann [436]
I dont think the answer is 129 unless you typed the problem in wrong. Order of operations say to do everything inside the perenthesis first so you would do the 3*2 which is 6 then divide the 12 by 6 and you get 2. so your new equation is 8^2+9(2)-7. Next distribute the 9 to the 2 that is inside the perenthesis. Square root the 8 also. Now it is 49+18-7. The answer would be 60 ...?

6 0
3 years ago
Three vertices of a rectangle are (–2, 1), (–2, –3), and (4, –3)
Tpy6a [65]
The answer is C
it's easier if you graph it
we know that the x of the unknown vertice is 4, because this is a rectangle and that vertices has to be on the same x line as the point below.
y of the unknown vertice has to be 1 because the vertice to the left needs to be on the same y line as the unknown vertice

6 0
3 years ago
A phone manufacturer wants to compete in the touch screen phone market. Management understands that the leading product has a le
CaHeK987 [17]

Answer:

Step-by-step explanation:

Hello!

To compete in the touch screen phone market a manufacturer aims to release a new touch screen with a battery life said to last more than two hours longer than the leading product which is the desired feature in phones.

To test this claim two samples were taken:

Sample 1

X: battery lifespan of a unit of the new product (min)

n= 93 units of the new product

mean battery life X[bar]= 8:53hs= 533min

S= 84 min

Sample 2

X: battery lifespan of a unit of the leading product (min)

n= 102 units of the leading product

mean battery life X[bar]= 5:40 hs = 340min

S= 93 min

The population variances of both variances are unknown and distinct.

To test if the average battery life of the new product is greater than the average battery life of the leading product by 2 hs (or 120 min) the parameters of interest will be the two population means and we will test their difference, the hypotheses are:

H₀: μ₁ - μ₂ ≤ 120

H₁:  μ₁ - μ₂ > 120

Considering that there is not enough information about the distribution of both variables, but both samples are big enough, we can apply the central limit theorem and approximate the distribution of both sample means to normal, this way we can use the standard normal:

Z= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2)}{\sqrt{\frac{S_1^2}{n_1} +\frac{S_2^2}{n_2}  } }

Z≈N(0;1)

Z= \frac{(533-340)-120}{\sqrt{\frac{84^2}{56} +\frac{93^2}{102}  } }= 5.028

I hope this helps!

3 0
4 years ago
Can someone help me with this question
bija089 [108]

Answer:

what is the lesson? abot?

Step-by-step explanation:

8 0
2 years ago
I need help bad my teacher do not teach me this
Stolb23 [73]

Answer:

i will help you

Step-by-step explanation:

do you have a picture of the assignment? or can you post a picture of it?

8 0
3 years ago
Read 2 more answers
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