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AnnZ [28]
3 years ago
12

A soup kettle has a capacity of 4.5 l. what is the kettle’s capacity in milliliters?(1 l = 1,000 ml)

Mathematics
1 answer:
wariber [46]3 years ago
8 0
The answer is 450ml so hope this helps all of you who had this problem
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A number to the 8th power divided by the same number to the 5th power equals 27. What is the number?
Yuliya22 [10]

x^8/x^5=27

x=3 when you flip the equation (rounded number)


7 0
3 years ago
Read 2 more answers
Radioactive fallout from testing atomic bombs drifted across a region. There were 220 people in the region at the time and 36 of
vazorg [7]

Answer:

(a) Yes, the death rate observed in the group unusually​ high.

(b) Yes, this prove that exposure to radiation increases the risk of​ cancer.

Step-by-step explanation:

In a region where radioactive fallout from testing atomic bombs drifted across, 36 of the 220 people died of cancer.

The sample proportion of the number of people dying of cancer in this region is:

\hat p =\frac{36}{220} =0.164

It is estimated by the cancer expert that there will be 33 cancer deaths in a group of this size.

The population proportion is: p=\frac{33}{220} =0.15

A hypothesis test for single proportion can be performed to test if the proportion of cancer deaths was high or not and whether it was due to the increase in radiation.

The hypothesis can be defined as:

<em>H</em>₀: The proportion of death due to cancer in this region is 0.15, i.e. <em>p</em> = 0.15.

<em>H</em>ₐ: The proportion of death due to cancer in this region is more than 0.15, i.e. <em>p</em> > 0.15.

Assuming the population is Normally distributed since the sample size is large.

Assume the level of significance is <em>α</em> = 0.05.

The test statistic is:

z=\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n}}}\\= \frac{0.164-0.15}{\sqrt{\frac{0.15(1-0.15)}{220}}} \\=0.581

The test statistic value is 0.581.

<u>Decision rule:</u>

If the <em>p</em>-value of the test is less than the significance level then the null hypothesis is rejected and vice versa.

The <em>p</em>-value of the test is:

P(Z>0.581)=1-P(Z

**Use a <em>z</em>-table for the <em>p</em>-value.

The <em>p-</em>value = 0.281 > <em>α</em> = 0.05.

The null hypothesis was failed to be rejected.

(a)

As the null hypothesis was not rejected at 5% level of significance it can be concluded that the death rate due to cancer observed in this group of people is unusually high.

Yes, the death rate observed in the group unusually​ high.

(b)

It was previously mentioned that a radioactive fallout from testing atomic bombs drifted across this region. Due to this 36 people died of cancer in this region.

As the death rate observed in the group unusually​ high, it can be said that the increase in the death rate due to cancer was due to the excessive exposure to radiation.

Yes, this prove that exposure to radiation increases the risk of​ cancer.

5 0
3 years ago
Which identity could be used to rewrite the expression x9-8?
RideAnS [48]

Answer:

C difference of squares

8 0
3 years ago
What is the reflection image of P (0, 0) after two reflections, first across x = −3 and then across y = −3? (−6, −6) (−6, −3) (−
Nadya [2.5K]

Answer:

(-3,-3)

Step-by-step explanation:

If we reflect P (0,0) across x=-3 and then across y=-3, then the new point of P would be (0-3,0-3)=(-3,-3)

4 0
4 years ago
Please help! I need these done before 2 pm!
jeyben [28]

Answer:

9. (7a + 6b – 9c) – (3a – 6c)

=7a+6b-9c-3a+6c

=7a-3a+6b-9c+6c

=4a+6b-3c

10. (x2 – 9) – (-2x2 + 5x – 3)

= x^2-9+2x^2-5x+3

=x^2+2x^2-5x-9+3  

=3x^2-5x-6

11. (5 – 6d – d2) – (-4d – d2)

=5-6d- d^2+4d+ d^2

=5-6d+4d-d^2+ d^2

=5-2d  

12. (-4x + 7) – (3x – 7)

=-4x+7-3x+7

= -4x-3x+7+7

=-7x+14

13. (4a – 3b) – (5a – 2b)

=4a-3b-5a+2b

=4a-5a-3b+2b

= -a-b

14. (2c + 3d) – (-6d – 5c)

=2c+3d+6d+5c

=2c+5c+3d+6d

=7c+9d

15. (5x2 + 6x – 9) – (x2 – 3x +7)

=5x^2+6x-9- x^2+3x-7

=5x^2- x^2+6x+3x-9-7

=4x^2+9x-16

16. (3y – 6) – (8 – 9y)

=3y-6-8+9y

=3y+9y-6-8

=12y-14

17. (3a2 – 2ab + 3b2) - (-a2 – 5ab + 3b2)

=3a^2-2ab+3b^2+ a^2+5ab-3b^2

=3a^2+ a^2-2ab+5ab+3b^2- 3b^2

=4a^2+3ab  

18. 5c – [8c – (6 – 3c)]

=5c-[8c-6+3c]

=5c-8c+6-3c

=5c-8c-3c+6

= -6c+6

19. 10x + [3x – (5x – 4)]

=10x+[ 3x-5x+4]

=10x+3x-5x+4

=8x+4

20. 3x 2 – [7x- (4x – x2) + 3]

=3x^2-[7x-4x+ x^2+3]

=3x^2-7x+4x- x^2-3

=3x^2-x^2-7x+4x-3

=2x^2-3x-3

21. x2 – [ - 3x+ ( 4 – 7x)]

= x^2-[ -3x+4-7x]

= x^2+3x-4+7x

= x^2+3x+7x-4

= x^2+10x-4

7 0
3 years ago
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