All categories

2 years ago

Point M is the midpoint of point A and point B. Find the coordinates of point B .

Mathematics

1 answer:

kkurt [141]2 years ago

8 0

Answer:

coordinate of point b is 12,1 by midpoint formula

Step-by-step explanation:

take mp formula and evaluate by putting value it will give you x again same process then it will give you u

You might be interested in

Soft drink consumption in New Zealand. A survey commissioned by the Southern Cross Healthcare Group reported that 16% of New Zea

lidiya [134]

<h2>Answer with explanation:</h2>

Given : The proportion of New Zealanders consume five or more servings of soft drinks per week : $\hat{p}= 0.16$

a) The number of survey respondents reported that they consume five or more servings of soft drinks per week = 2006

b) Confidence interval for population proportion (p) :

$\hat{p}\pm z^*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

, where $\hat{p}$ = Sample proportion.

n= Sample size.

z* = Critical value.

For n= 2006 , $\hat{p}= 0.16$ and critical value for 95% confidence interval : z* = 1.96

Then , the required confidence interval will be :

$0.16\pm (1.96)\sqrt{\dfrac{ 0.16(1- 0.16)}{2006}}$

$0.16\pm (1.96)\sqrt{ 0.000066999002991}$

$0.16\pm (1.96)(0.00818529186963)$

$0.16\pm 0.0160431720645$

$(0.16-0.0160431720645,\ 0.16+0.0160431720645)$

$(0.143956827935,\ 0.176043172064)\approx(0.144,\ 0.176)$

i.e. A 95% confidence interval for the proportion of New Zealanders who report that they consume five or more servings of soft drinks per week. = $(0.144,\ 0.176)$

c) $(0.144,\ 0.176)=(0.144\times100\%,\ 0.176\times100\%)$

$=(14.4\%,\ 17.6\%)$ 176\times100\%)[/tex]

d) The estimate might be biased because the survey is taken online that means offline New Zealanders are out of consideration.

6 0

2 years ago

Consider an urn containing 8 white balls, 7 red balls and 5 black balls.

weqwewe [10]

Answer + Step-by-step explanation:

1) The probability of getting 2 white balls is equal to:

$=\frac{8}{20} \times \frac{7}{19}\\\\= 0.147368421053$

2) the probability of getting 2 white balls is equal to:

$=C^{2}_{5}\times (\frac{8}{20} \times \frac{7}{19}) \times (\frac{12}{18} \times \frac{11}{17} \times \frac{10}{16})\\=0.397316821465$

3) The probability of getting at least 72 white balls is:

$=C^{72}_{150}\times \left( \frac{8}{20} \right)^{72} \times \left( \frac{7}{20} \right)^{78} +C^{73}_{150}\times \left( \frac{8}{20} \right)^{73} \times \left( \frac{7}{20} \right)^{77} + \cdots +C^{149}_{150}\times \left( \frac{8}{20} \right)^{149} \times \left( \frac{7}{20} \right)^{1} +\left( \frac{8}{20} \right)^{150}$