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Y_Kistochka [10]
4 years ago
15

Evaluate 6+3b if b=7

Mathematics
2 answers:
Pavlova-9 [17]4 years ago
6 0
27. Replace b with 7, making 6 + 3(7). The 3 and parentheses of 7 hint multiplication, so you multiply to make 21, then add 6 relating back to PEMDAS to teach you the correct order to solve the problem.
Wittaler [7]4 years ago
5 0
Evaluate 6+3b if b=7
To solve this problem, we must plug the value of b into the equation and follow the order of operations like so:
Original Equation: 6+3b
Plug in Given Information: 6+3*7
Remember that in the order of operations, we multiply before adding.
Multiply: 6+21
Add: The answer is <u>27.</u>

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A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 400 babies were​ born, a
Masja [62]

Answer:

(a) 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) Yes​, the proportion of girls is significantly different from 0.50.

Step-by-step explanation:

We are given that a clinical trial tests a method designed to increase the probability of conceiving a girl.

In the study 400 babies were​ born, and 340 of them were girls.

(a) Firstly, the pivotal quantity for 99% confidence interval for the population 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 girls born = \frac{340}{400} = 0.85

             n = sample of babies = 400

             p = population percentage of girls born

<em>Here for constructing 99% confidence interval we have used One-sample z proportion statistics.</em>

<u>So, 99% confidence interval for the population proportion, p is ;</u>

P(-2.58 < N(0,1) < 2.58) = 0.99  {As the critical value of z at 0.5% level

                                                    of significance are -2.58 & 2.58}  

P(-2.58 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 2.58) = 0.99

P( -2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

P( \hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.99

<u>99% confidence interval for p</u> = [\hat p-2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+2.58 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

= [ 0.85-2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } } , 0.85+2.58 \times {\sqrt{\frac{0.85(1-0.85)}{400} } } ]

 = [0.804 , 0.896]

Therefore, 99% confidence interval for the percentage of girls born is [0.804 , 0.896].

(b) <em>Let p = population proportion of girls born.</em>

So, Null Hypothesis, H_0 : p = 0.50      {means that the proportion of girls is equal to 0.50}

Alternate Hypothesis, H_A : p \neq 0.50      {means that the proportion of girls is significantly different from 0.50}

The test statistics that will be used here is <u>One-sample z proportion test</u> <u>statistics</u>;

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

where, \hat p = sample proportion of girls born = \frac{340}{400} = 0.85

             n = sample of babies = 400

So, <u><em>the test statistics</em></u>  =  \frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }

                                     =  19.604

Now, at 0.01 significance level, the z table gives critical value of 2.3263 for right tailed test. Since our test statistics is way more than the critical value of z as 19.604 > 2.3263, so we have sufficient evidence to reject our null hypothesis due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the proportion of girls is significantly different from 0.50.

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In this unit, you calculated the surface area of solid figures composed of polygons, such as rectangles and triangles. But imagi
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Answer:I don't know the exact problem but I'll just give you the formulas on how to find volume, circumference radious, and yada

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Answer:

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If you observe the 4 graphical representations you will see that except stem and leaf plot, the rest 3 represent the same information. 

As the time is increasing, the frequency of students who listen to music is decreasing. The frequency is maximum for the first group 0 - 1 and least for the last group 6 - 7.

The ranges shown in both Frequency Table and Histogram are the same. 

So, the following options are correct:
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