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The magnitude of vector a = (5,m) is 13 and the magnitude of vector b = (n, 24) is 25. What are m and n

Ulleksa [173]

Answer:

$m=12\,,n=7$

Step-by-step explanation:

The magnitude of vector $(x,y)$ is given by $\sqrt{x^2+y^2}$

The magnitude of vector $a=(5,m)$ is $13$ .

$\sqrt{5^2+m^2}=13\\5^2+m^2=13^2\\25+m^2=169\\m^2=169-25\\m^2=144\\m=12$

The magnitude of vector $b=(n,24)$ is $25$ .

$\sqrt{n^2+24^2}=25\\n^2+24^2=25^2\\n^2+576=625\\n^2=625-576\\n^2=49\\n=7$

Therefore,

$m=12\,,n=7$

5 0

3 years ago

Answer right you get brainiest also 20 points just number 1 and 2

Klio2033 [76]

Answer:

1. Newton's law of gravitation. Newton's law of gravitation, statement that any particle of matter in the universe attracts any other with a force varying directly as the product of the masses and inversely as the square of the distance between them.

2. The key difference is that Newtonian gravity has a privileged separation of spacetime into space and time, whereas Einsteinian gravity just has spacetime. Edit: to be absolutely clear, Newtonian gravity can be written as spacetime curvature! This is counter to the common statements about the novel thing in GR.

Step-by-step explanation:

4 0

3 years ago

Read 2 more answers

If you toss two pennies what are the chances of getting two heads?

Norma-Jean [14]

Answer:

50/50 chance i guessing

If theres 2 pennies

<h2>Hope this helps :)</h2>

4 0

3 years ago

32.48x12.56 please help my day show her work

Alekssandra [29.7K]

The answer would be 4,060,952

7 0

3 years ago

Consider a random sample of ten children selected from a population of infants receiving antacids that contain aluminum, in orde

saveliy_v [14]

Answer:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids.

b. (32.1, 42.3)

c. p-value < .00001

d. The null hypothesis is rejected at the α=0.05 significance level

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

p-value equals < .00001. The null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids<em> </em>greatly increases the plasma aluminum levels of children.

Step-by-step explanation:

a. Null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is different from the mean plasma aluminum level of the population of infants not receiving antacids. This may imply that being given antacids significantly changes the plasma aluminum level of infants.

b. Since the population standard deviation σ is unknown, we must use the t distribution to find 95% confidence limits for μ. For a t distribution with 10-1=9 degrees of freedom, 95% of the observations lie between -2.262 and 2.262. Therefore, replacing σ with s, a 95% confidence interval for the population mean μ is:

(X bar - 2.262\frac{s}{\sqrt{10} } , X bar + 2.262\frac{s}{\sqrt{10} })

Substituting in the values of X bar and s, the interval becomes:

(37.2 - 2.262\frac{7.13}{\sqrt{10} } , 37.2 + 2.262\frac{7.13}{\sqrt{10} })

or (32.1, 42.3)

c. To calculate p-value of the sample , we need to calculate the t-statistics which equals:

t=\frac{(Xbar-u)}{\frac{s}{\sqrt{10} } } = \frac{(37.2-4.13)}{\frac{7.13}{\sqrt{10} } } = 14.67.

Given two-sided test and degrees of freedom = 9, the p-value equals < .00001, which is less than 0.05.

d. The mean plasma aluminum level for the population of infants not receiving antacids is 4.13 ug/l - not a plausible value of mean plasma aluminum level for the population of infants receiving antacids. The 95% confidence interval for the population mean of infants receiving antacids is (32.1, 42.3) and does not cover the value 4.13. Therefore, the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids <em>greatly changes</em> the plasma aluminum levels of children.

e. Reformulated null hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is equal to the mean plasma aluminum level of the population of infants not receiving antacids.

Reformulated complementary alternative hypothesis: The mean plasma aluminum level of the population of infants receiving antacids is higher than the mean plasma aluminum level of the population of infants not receiving antacids.

Given one-sided test and degree of freedom = 9, the p-value equals < .00001, which is less than 0.05. This result is similar to result in part (c). the null hypothesis is rejected at the α=0.05 significance level. This suggests that being given antacids<em> greatly increases</em> the plasma aluminum levels of children.

6 0

3 years ago