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Irina-Kira [14]

3 years ago

5

Can someone do the division and show work of 350000 divided by 35

1 answer:

Digiron [165]3 years ago

7 0

Answer:

10000

Step-by-step explanation:

I hope this helps!

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What is the first quartile of the box-and-whisker plot?

iogann1982 [59]

B) 48 is your answer

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4 years ago

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Write an equation relating c, the cost, and p, the pounds of strawberries.

pav-90 [236]

Answer: p

c

Step-by-step explanation:

5 0

4 years ago

In this exercise, T: R2 → R2 is a function. For each of the following parts, state why T is not linear. (a) T(a1, a2)= (1, a2) (

DENIUS [597]

Answer:

a) T is not homogeneous
b) T is not additive
c) T is not homogeneous
d) T is not additive
e) T is not additive

Step-by-step explanation:

In each case there is an example where one property of linear functions fails.

a) $T(2(1,0))=T((2,0))=(1,0); 2T((1,0))=2(1,0)=(2,0)$ . These vectors are not equal, then T doesn't satisfy the condition of scalar multiplication (homogeneity).

b) $T((1,2)+(2,3))=T(3,5)=(3,9); T((1,2))+T((2,3))=(1,1)+(2,4)=(3,5)$ . Because these vectors are not equal, T doesn't satisfy the property of vector addition (additivity).

c) $T(\frac{1}{2}(\pi,0))=T((\frac{\pi}{2},0))=(\sin(\frac{\pi}{2}),0)=(1,0); \frac{1}{2}T((\pi,0))=\frac{1}{2}(0,0)=(0,0)$ so T is not homogeneous.

d) $T((-1,0)+(1,0))=T(0,0)=(0,0); T((-1,0))+T((1,0))=(1,0)+(1,0)=(2,0)$ then T is not additive.

e) $T((0,0)+(1,0))=T(1,0)=(2,0); T((0,0))+T((1,0))=(1,0)+(2,0)=(3,0)$ then T is not additive.

8 0

3 years ago

A manufacturer of college textbooks is interested in estimating the strength of the bindings produced by a particular binding ma

IrinaVladis [17]

Answer:

538 books should be tested.

Step-by-step explanation:

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1-0.99}{2} = 0.005$

Now, we have to find z in the Ztable as such z has a pvalue of $1-\alpha$ .

So it is z with a pvalue of $1-0.005 = 0.995$ , so $z = 2.575$

Now, find the margin of error M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

How many books should be tested to estimate the average force required to break the binding to within 0.08 lb with 99% confidence?

n books should be tested.

n is found when $M = 0.08$

We have that $\sigma = 0.72$

$M = z*\frac{\sigma}{\sqrt{n}}$

$0.08 = 2.575*\frac{0.72}{\sqrt{n}}$

$0.08\sqrt{n} = 2.575*0.72$

$\sqrt{n} = \frac{2.575*0.72}{0.08}$

$(\sqrt{n})^{2} = (\frac{2.575*0.72}{0.08})^{2}$

$n = 537.1$

Rounding up

538 books should be tested.

6 0

3 years ago

Help me please show your work I’ll give brainlest.

serious [3.7K]

1. 0.8
2. 3.87298
3. ?
4. ?
5. ?
6. 5
7. True
8. ?
9.

3 0

3 years ago