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3 years ago

What is the value of the 9 in 913256

Mathematics

2 answers:

emmainna [20.7K]3 years ago

5 0

Hundred thousands// 900,000

Masja [62]3 years ago

4 0

Well ther is place values 1 ones,2 tenths,3 hundreds,4 thousands,5 ten thousands,6 hundred thousands,7 million. so take that and enpliy the places so the answer will be hundred thousands

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Find the number of sides for a regular polygon whose interior angles each measure 10 times each exterior angle

gavmur [86]

Answer:

22 sides

Step-by-step explanation:

The expression to find an interior angle of a polygon is:

$\frac{(n-2)*180}{n}$

The expression to find an exterior angle of a polygon is:

$\frac{360}{n}$

Please note that "n" represents the number of sides the polygon has.

We can use these two expressions to set up an equation.

$\frac{(n-2)*180}{n}=10(\frac{360}{n})$

Multiply both sides by "n":

$(n-2)*180=10n(\frac{360}{n})$

Now, distribute:

$180(n)-180(2)=\frac{3600n}{n}\\180n-360=3600$

Divide both sides by 10:

$\frac{180n}{10}-\frac{360}{10}=\frac{3600}{10}\\$

$18n-36=360$

Add 36 to both sides:

$18n=360+36\\18n=396$

Divide both sides by 18:

$n=22$

The polygon has 22 sides

6 0

3 years ago

In a sample of nequals16 lichen specimens, the researchers found the mean and standard deviation of the amount of the radioacti

GuDViN [60]

Answer:

(a) The confidence level desired by the researchers is 95%.

(b) The sampling error is 0.002 microcurie per millilitre.

Step-by-step explanation:

The mean and standard deviation of the amount of the radioactive element, cesium-137 present in a sample of n = 16 lichen specimen are:

$\bar x=0.009\\s=0.005$

Now it is provided that the researchers want to increase the sample size in order to estimate the mean μ to within 0.002 microcurie per millilitre of its true value, using a 95% confidence interval.

The (1 - α)% confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

(a)

The confidence level is the probability that a particular value of the parameter under study falls within a specific interval of values.

In this case the researches wants to estimate the mean using the 95% confidence interval.

Thus, the confidence level desired by the researchers is 95%.

(b)

In case of statistical analysis, during the computation of a certain statistic, to estimate the value of the parameter under study, certain error occurs which are known as the sampling error.

In case of the estimate of parameter using a confidence interval the sampling error is known as the margin of error.

In this case the margin of error is 0.002 microcurie per millilitre.

(c)

The margin of error is computed using the formula:

$MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

The critical value of z for 95% confidence level is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

*Use a z-table.

$MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}$

$0.002=1.96\times \frac{0.005}{\sqrt{n}}$

$n=[\frac{1.96\times 0.005}{0.002}]^{2}$

$=(4.9)^{2}\\=24.01\\\approx 25$

Thus, the sample size necessary to obtain the desired estimate is 25.

6 0

4 years ago

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

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

So, 99% confidence interval for the population proportion, p is ;

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

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

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

= [0.804 , 0.896]

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

(b) Let p = population proportion of girls born.

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 One-sample z proportion test statistics;

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, the test statistics = $\frac{0.85-0.50}{\sqrt{\frac{0.85(1-0.85)}{400} } }$

= 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 we reject our null hypothesis.

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

8 0

3 years ago

If you hike three miles every 5 minutes, how far are you going in 1 minute?

V125BC [204]

Answer:

0.6 miles per minute

Step-by-step explanation:

3/5 = 0.6/1 = 0.6

6 0

3 years ago

Read 2 more answers

A restaurant owner bought b large bags of flour for $45 each and b large bags of sugar for $25 each. The expression b × 45 + b ×

sammy [17]

45b+25b :) please mark me brainliest. have a great day!!

5 0

3 years ago