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3 years ago

Katie has read 32 % of a book. If she has read 80 pages, how many more pages does Katie have left to read?

Mathematics

2 answers:

gizmo_the_mogwai [7]3 years ago

8 0

Answer:

it is 170 pages

Step-by-step explanation:

you change 32% to a fraction 32/100 then figure the rest out cause the rest of it is mental math.

mestny [16]3 years ago

7 0

Percentage of pages of a book read by Katie = 32%
Total number of pages of the book read by Katie = 80
Let us assume the number of pages that the book has = x
Then
(32/100) * x = 80
32x = 80 * 100
32x = 8000
x = 8000/32
= 250
So the book has a total of 250 pages in it.
Then
The number of pages that Katie still needs to read = 250 - 80
= 170
So Katie needs to read 170 more pages in the book.

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1 year ago

The pesticide diazinon is in common use to treat infestations of the German cockroach, Blattella germanica. A study investigated

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We conclude that there is no difference in the proportion of cockroaches that died on each surface.

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We are given that a study investigated the persistence of this pesticide on various types of surfaces.

After 14 days, they randomly assigned 72 cockroaches to two groups of 36, placed one group on each surface, and recorded the number that died within 48 hours. On the glass, 18 cockroaches died, while on plasterboard, 25 died.

Let $p_1$ = proportion of cockroaches that died on glass surface.

$p_2$ = proportion of cockroaches that died on plasterboard surface.

So, Null Hypothesis, $H_0$ : $p_1-p_2$ = 0 {means that there is no difference in the proportion of cockroaches that died on each surface}

Alternate Hypothesis, $H_A$ : $p_1-p_2\neq$ 0 {means that there is a significant difference in the proportion of cockroaches that died on each surface}

The test statistics that would be used here Two-sample z proportion statistics;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of cockroaches that died on glass surface = $\frac{18}{36}$ = 0.50

$\hat p_2$ = sample proportion of cockroaches that died on plasterboard surface = $\frac{25}{36}$ = 0.694

$n_1$ = sample of cockroaches on glass surface = 36

$n_2$ = sample of cockroaches on plasterboard surface = 36

So, test statistics = $\frac{(0.50-0.694)-(0)}{\sqrt{\frac{0.50(1-0.50)}{36}+\frac{0.694(1-0.694)}{36} } }$

= -1.712

The value of z test statistics is -1.712.

Since, in the question we are not given with the level of significance so we assume it to be 5%. Now, at 5% significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistics lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the proportion of cockroaches that died on each surface.

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