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larisa86 [58]
3 years ago
13

The size of a large milkshake is 1.4 times the size of a medium milkshake. Write 1.4 as a percent

Mathematics
2 answers:
pishuonlain [190]3 years ago
6 0
25%
---------------------------------
Lapatulllka [165]3 years ago
4 0
140% i believe. I'm not 100% sure

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A doctor released the results of clinical trials for a vaccine to prevent a particular disease. In these clinical​ trials, 400 c
Soloha48 [4]

Answer:

We conclude that the vaccine appears to be​ effective.

Step-by-step explanation:

We are given that a doctor released the results of clinical trials for a vaccine to prevent a particular disease.

The subjects in group 1​ (the experimental​ group) were given the​ vaccine, while the subjects in group 2​ (the control​ group) were given a placebo. Of the 200 comma 000 children in the experimental​ group, 38 developed the disease. Of the 200 comma 000 children in the control​ group, 81 developed the disease.

Let<em> </em>p_1<em> = proportion of subjects in the experimental​ group who developed the disease.</em>

<em />p_2<em> = proportion of subjects in the control​ group who developed the disease.</em>

So, Null Hypothesis, H_0 : p_1\geq p_2       {means that the vaccine does not appears to be​ effective}

Alternate Hypothesis, H_A : p_1       {means that the vaccine appears to be​ effective}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       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 children in the experimental​ group who developed the disease = \frac{38}{200,000} =  0.00019

\hat p_2 = sample proportion of children in the control​ group who developed the disease = \frac{81}{200,000} =  0.00041

n_1 = sample of children in the experimental​ group = 200,000

n_2 = sample of children in the control​ group = 200,000

So, <u><em>test statistics</em></u>  =  \frac{(0.00019-0.00041)-(0)}{\sqrt{\frac{0.00019(1-0.00019)}{200,000}+\frac{0.00041(1-0.00041)}{200,000}  } }

                              =  -4.02

The value of z test statistics is -4.02.

Now, at 0.01 significance level the z table gives critical values of -2.33 for left-tailed test.

Since our test statistics is less than the critical value of z as -4.02 < -2.33, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the vaccine appears to be​ effective.

3 0
3 years ago
What is the sum of 10/12and 11/12? Make sure to show all work to receive full credit!
sergiy2304 [10]

Answer:

1\frac{3}{4}

Step-by-step explanation:

\frac{10}{12} +\frac{11}{12} \\\\\frac{21}{12} \\\\1\frac{9}{12} \\\\1\frac{3}{4}

7 0
3 years ago
Electric charge is distributed over the disk x2 + y2 ≤ 16 so that the charge density at (x, y) is rho(x, y) = 2x + 2y + 2x2 + 2y
professor190 [17]

Answer:

Required total charge is 256\pi coulombs per square meter.

Step-by-step explanation:

Given electric charge is dristributed over the disk,

x^2=y^2\leq 16 so that the charge density at (x,y) is,

\rho (x,y)=2x+2y+2x^2+2y^2

To find total charge on the disk let Q be the total charge and x=r\cos\theta,y=r\sin\theta so that,

Q={\int\int}_Q\rho(x,y) dA                where A is the surface of disk.

=\int_{0}^{2\pi}\int_{0}^{4}(2x+2y+2x^2+2y^2)dA

=\int_{0}^{2\pi}\int_{0}^{4}(2r\cos\theta+2r\sin\theta+2r^2 \cos^{2}\theta+2r^2\sin^2\theta)rdrd\theta

=2\int_{0}^{2\pi}\int_{0}^{4}r^2(\cos\theta+\sin\theta)drd\theta+2\int_{0}^{2\pi}\int_{0}^{4}r^3drd\theta

=\frac{2}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)\Big[r^3\Big]_{0}^{4}d\theta+2\int_{0}^{2\pi}\Big[\frac{r^4}{4}\Big]d\theta

=\frac{128}{3}\int_{0}^{2\pi}(\sin\theta+\cos\theta)d\theta+128\int_{0}^{2\pi}d\theta

=\frac{128}{3}\Big[\sin\theta-\cos\theta\Big]_{0}^{2\pi}+128\times 2\pi

=\frac{128}{3}\Big[\sin 2\pi-\cos 2\pi-\sin 0+\cos 0\Big]+256\pi

=256\pi

Hence total charge is 256\pi coulombs per square meter.

3 0
3 years ago
Mexico City, Mexico, is the world's second largest metropolis and is also one of its fastest-growing cities with a projected gro
Cloud [144]

Answer:

Predicted population of Mexico City in year 2010 = 38386000

Step-by-step explanation:

Formula to be used for the population of Mexico,

P = 20.899e^{0.032t}

Where t = Duration after year 1991

P = Final population

Number of years between 2010 and 1991 = 19 years

Predicted population of Mexico city after 2010 = 20.899e^{0.032\times 19}

                                                                              = 20.899 × (1.8368)

                                                                              = 38.38633 million

                                                                              = 38.386 million

Therefore, predicted population of Mexico City in year 2010 = 38386000

5 0
3 years ago
In statistics, the middle value of an ordered set of values is called what?
zaharov [31]
This is called the median.
8 0
3 years ago
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