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dybincka [34]
1 year ago
5

Tiana has already taken 1 page of notes on her own, and she will take 1 page during each hour of class. In all, how many hours w

ill Tiana have to spend in class before she will have a total of 43 pages of notes in her notebook?
Mathematics
1 answer:
Novosadov [1.4K]1 year ago
4 0

Answer:

<u>42 Hours.</u>

Step-by-step explanation:

As an equation where y is the total pages and x is hours of class:

y=1+1x since she started with 1 page and does 1 page per hour.

Setting the total pages she needs as 43 (y=43) we get:

43=1+1x

Subtract one on both sides

42=1x

The 1 doesn't need to be written so

x=42

Therefore, it'll take her 42 hours to write all those notes.

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Body armor provides critical protection for law enforcement personnel, but it does affect balance and mobility. The article "Imp
yKpoI14uk [10]

Answer:

No, there is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

Step-by-step explanation:

This is a hypothesis test for the population mean.

The claim is that true average task time with armor is less than 2 seconds.

Then, the null and alternative hypothesis are:

H_0: \mu=2\\\\H_a:\mu< 2

The significance level is 0.01.

The sample has a size n=52.

The sample mean is M=1.95.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=0.2.

The estimated standard error of the mean is computed using the formula:

s_M=\dfrac{s}{\sqrt{n}}=\dfrac{0.2}{\sqrt{52}}=0.028

Then, we can calculate the t-statistic as:

t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{1.95-2}{0.028}=\dfrac{-0.05}{0.028}=-1.803

The degrees of freedom for this sample size are:

df=n-1=52-1=51

This test is a left-tailed test, with 51 degrees of freedom and t=-1.803, so the P-value for this test is calculated as (using a t-table):

P-value=P(t

As the P-value (0.039) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that true average task time with armor is less than 2 seconds.

3 0
1 year ago
Kendra has a bowl containing 10 blue marbles, 15 yellow marbles, 5 green marbles, and 6 red marbles of the same size. She will r
Mice21 [21]
Given:
10 blue marbles ; 15 yellow marbles, 5 green marbles, 6 red marbles
A total of 36 marbles.

Probabilities:
1st marble: red = 6/36
2nd marble: yellow = 15/35

6/36 * 15/35 = 6*15 / 36*35 = 90/1260 = 1/14

The probability that Kendra will draw a red marble and then a yellow marble is 1/14.
6 0
2 years ago
A pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome). In the first sta
Akimi4 [234]

Answer:

95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

Step-by-step explanation:

We are given that a pharmaceutical company proposes a new drug treatment for alleviating symptoms of PMS (premenstrual syndrome).

In the first stages of a clinical trial, it was successful for 7 out of the 14 women.

Firstly, the pivotal quantity for 95% confidence interval for the true proportion is given by;

                             P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample proportion of women who find success with this new treatment = \frac{7}{14} = 0.50

          n = sample of women = 14

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

So, 95% confidence interval for the true proportion, p is ;

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5%

                                            level of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u />

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.50-1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } , 0.50+1.96 \times {\sqrt{\frac{0.50(1-0.50)}{14} } } ]

 = [0.238 , 0.762]

Therefore, 95% confidence interval for p, the true proportion of all women who will find success with this new treatment is [0.238 , 0.762].

4 0
2 years ago
Emily an her dad share the cost of the meal in the ratio 2:3 emily’s Dad pays £52.20 what is the total cost of the meal
rodikova [14]

£87.00

Emily's dad pays 3 parts of the meal

Divide £52.20 by 3 to find one part of the ratio

\frac{52.20}{3} = £17.40 ← 1 part of the ratio

2 parts = 2 × £17.40 = £34.80 ← Emily's share

total cost = £52.20 + £34.80 = £87.00


6 0
2 years ago
Read 2 more answers
Mike is 5 years old and Randy is 8 years old. Select all proportional relationships that have the same ratio to that of Mike and
Yanka [14]

Answer:

The ratio is multiplying 1.6 so just pick all the proportional relationships where you have to multiply by 1.6.

Step-by-step explanation:

Proportional relationships man you have to multiply, but since I didn't know what you multiplied 5 to get to 8 I divided.

8÷5=1.6

And if you want to check your answer then multiply.

5×1.6=8

<u><em>Could I please have BRAINLIEST?</em></u>

5 0
2 years ago
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