What is 2,789 to the nearest ten

Suppose that the Celsius temperature at the point (x, y) in the xy-plane is T(x, y) = x sin 2y and that distance in the xy-plane

liraira [26]

Missing information:

How fast is the temperature experienced by the particle changing in degrees Celsius per meter at the point

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

Answer:

$Rate = 0.935042^\circ /cm$

Step-by-step explanation:

Given

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$T(x,y) =x\sin2y$

$r = 1m$

$v = 2m/s$

Express the given point P as a unit tangent vector:

$P = (\frac{1}{2}, \frac{\sqrt 3}{2})$

$u = \frac{\sqrt 3}{2}i - \frac{1}{2}j$

Next, find the gradient of P and T using: $\triangle T = \nabla T * u$

Where

$\nabla T|_{(\frac{1}{2}, \frac{\sqrt 3}{2})} = (sin \sqrt 3)i + (cos \sqrt 3)j$

So: the gradient becomes:

$\triangle T = \nabla T * u$

$\triangle T = [(sin \sqrt 3)i + (cos \sqrt 3)j] * [\frac{\sqrt 3}{2}i - \frac{1}{2}j]$

By vector multiplication, we have:

$\triangle T = (sin \sqrt 3)* \frac{\sqrt 3}{2} - (cos \sqrt 3) \frac{1}{2}$

$\triangle T = 0.9870 * 0.8660 - (-0.1606 * 0.5)$

$\triangle T = 0.9870 * 0.8660 +0.1606 * 0.5$

$\triangle T = 0.935042$

Hence, the rate is:

$Rate = \triangle T = 0.935042^\circ /cm$

3 0

2 years ago

X-1.6=2.4+0.8x solve for x. What did I do wrong?

trapecia [35]

$\bf x-1.6=2.4+0.8x\implies \stackrel{\textit{subtracting 2.4}}{x-4=0.8x}\implies \stackrel{\textit{subtracting x}}{-4=0.8x-x} \\\\\\ -4=-0.2x\implies \cfrac{-4}{-0.2}=x\implies 20=x$

3 0

3 years ago

Manufacturers are testing a die to make sure that it is fair (has a uniform distribution). They roll the die 66 times and record

steposvetlana [31]

Answer:

We should reject H0

At the 5% significance level, there is sufficient evidence to conclude that the die is not fair.

Step-by-step explanation:

Null hypothesis is a statement that is to be tested against the alternative hypothesis and then decision is taken whether to accept or reject the null hypothesis. The critical value is 15.091 and test statistic is 11.070. The null hypothesis is rejected or accepted on the basis of level of significance. When the p-value Test statistics is greater than level of significance we fail to reject the null hypothesis and null hypothesis is then accepted. In the given case p-value is less than critical value then we should reject the null hypothesis.

6 0

2 years ago

GIVING BRAINLIEST AND THE REST OF MY POINTSSS!!!!!PLS HELPP :((

mezya [45]

Answer:

the bowling alley

Step-by-step explanation:

7 0

2 years ago

Read 2 more answers

A sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8. If the sample mean corre

jeka57 [31]

Answer:

The value of the sample mean is 78.

Step-by-step explanation:

We are given that a sample of n = 4 scores is obtained from a population with a mean of 70 and a standard deviation of 8.

Also, the sample mean corresponds to a z score of 2.00.

Let $\bar X$ = sample mean

The z-score probability distribution for a sample mean is given by;

Z = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\mu$ = population mean = 70

$\sigma$ = standard deviation = 8

n = sample size = 4

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, we are given that the sample mean corresponds to a z score of 2.00 for which we have to find the value of sample mean;

So, z-score formula is given by ;

z-score = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ = $\frac{\bar X-70}{\frac{8}{\sqrt{4} } }$

2.00 = $\frac{\bar X-70}{\frac{8}{\sqrt{4} } }$

2.00 = $\frac{\bar X-70}{4 } }$

$\bar X = 70+(2 \times 4)$

$\bar X$ = 70 + 8 = 78

Therefore, the value of the sample mean is 78.

3 0

2 years ago