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4 years ago

Task: E = mc2

Mathematics

1 answer:

Afina-wow [57]4 years ago

3 0

Answer:

$c=\left(\frac{E}{m} \right)^{\frac{1}{2} }$

Step-by-step explanation:

$E = mc^2$

$c^2=\frac{E}{m}$

$c=\sqrt{\frac{E}{m} }$

$c=\left(\frac{E}{m} \right)^{\frac{1}{2} }$

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Ten college students were randomly selected. Their grade point averages (GPAs) when they entered the program were between 3.5 a

solong [7]

Answer:

The 95% confidence interval for the true slope is (0.03985, 0.14206).

Step-by-step explanation:

For the regression equation:

$\hat y=\alpha +\hat \beta x$

The (1 - <em>α</em>)% confidence interval for the regression coefficient or slope $(\hat \beta )$ is:

$Ci=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )$

The regression equation for current GPA (Y) of students based on their GPA's when entering the program (X) is:

$\hat Y=3.584756+0.090953 X$

The summary of the regression analysis is:

Predictor Coefficient SE t-stat p-value

Constant 3.584756 0.078183 45.85075 5.66 x 10⁻¹¹

Entering GPA 0.090953 0.022162 4.103932 0.003419

The regression coefficient and standard error are:

$\hat \beta = 0.090953\\SE (\hat \beta)=0.022162$

The critical value of <em>t</em> for 95% confidence level and 8 degrees of freedom is:

$t_{\alpha/2, n-2}=t_{0.05/2, 10-2}=t_{0.025, 8}=2.306$

Compute the 95% confidence interval for $(\hat \beta )$ as follows:

$CI=\hat \beta \pm t_{\alpha/2, (n-2)}\times SE(\hat \beta )\\=0.090953\pm 2.306\times 0.022162\\=0.090953\pm 0.051105572\\=(0.039847428, 0.142058572)\\\approx (0.03985, 0.14206)$

Thus, the 95% confidence interval for the true slope is (0.03985, 0.14206).

6 0

3 years ago

What is the name of a figure made of 6 congruent squares

ozzi

Answer:

B) Cube

Step-by-step explanation:

A) square pyramid has one square base and 4 congruent triangles

B) Cube has 6 congruent squares

C) Rectangular pyramid has one rectangle base and 4 congruent triangles

D) Rectangular prism has 3 sets of rectangles; total of 6 rectangles

5 0

3 years ago

In an arithmetic sequence, the 10th term is 37, and the 21st term is 4. Find the common difference and the first term

Masja [62]

Answer:

Common difference is 3. And the first term is 64.

Step-by-step explanation:

The differnce between 21 and 10 is 11. So there are 11 unknown numbers in between. And the difference between 37 and 4 is 33. 33/11 is equal to 3. So the pattern is minus three. the first term is 9 terms before 37. 9 times 3 is 27. 37+27 is equal to 64. so the first term is 64

3 0

3 years ago

From experience, it is known that on average 10% of welds performed by a particular welder are defective. if this welder is requ

bulgar [2K]

Binomial distribution can be used because the situation satisfies all the following conditions:1. Number of trials is known and remains constant (n)2. Each trial is Bernoulli (i.e. exactly two possible outcomes) (success/failure)3. Probability is known and remains constant throughout the trials (p)4. All trials are random and independent of the othersThe number of successes, x, is then given by $P(x)=C(n,x)p^x(1-p)^{n-x}$ where $C(n,x)=\frac{n!}{x!(n-x)!}$
Here we're given
p=0.10 [ success = defective ]
n=3

(a) x=0
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,0)0.1^0(1-0.1)^{3-0}$
$=1(1)(0.729)$
$=0.729$

(b) x=2
$P(x)=C(n,x)p^x(1-p)^{n-x}$
$=C(3,2)0.1^2(1-0.1)^{3-2}$
$=3(0.01)(0.9)$
$=0.027$

(c) x ≥ 2
$P(x)=\sum_{x=2}^3C(n,x)p^x(1-p)^{n-x}$
$=P(2)+P(3)$
$=C(n,2)p^2(1-p)^{n-2}+C(n,3)p^3(1-p)^{n-3}$
$=C(3,2)0.1^2(1-0.1)^{3-2}+C(3,3)0.1^3(1-0.1)^{3-3}$
$=3(0.01)(0.9)+1(0.001)1$
$=0.027+0.001$
$=0.028$

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3 years ago

Choose the fraction that is equal to 0.245 A. 49/10 B. 245/999 C. 245/900 D. 245/1000

kicyunya [14]

D is equal to 0.245.

6 0

3 years ago