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3 years ago

I really need help i think my brain died

Mathematics

1 answer:

lyudmila [28]3 years ago

4 0

Same cause school kinda slow bsndndndndnndnfnfnfnfn sorry I need the points

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Please help find the value of y and the measures of two missing angles

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50 n 70 use thathopes it helps

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3 years ago

10^-3 mm in inches??

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it has to be 20 characters long so hi

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3 years ago

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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

2 years ago

Which pair of numbers is relatively prime?

Vladimir [108]

I think it is 21 and 50. The only common factor they have is 1.

All the others are divisible by 1 & 3.

3 0

3 years ago

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Please help me out with thissss

vova2212 [387]

Answer:

Step-by-step explanation:

From table 1,

f(x) = bˣ

For x = -1,

f(-1) = 0.5

0.5 = (b)⁻¹

b = $\frac{1}{0.5}$

b = 2

For x = 1.585,

f(1.585) = 3

3 = $2^{1.585}$

3 = $2^{1}\times2^{0.585}$

$2^{0.585}=\frac{3}{2}$

$2^{0.585}=1.5$

For x = 2.585,

f(2.585) = $2^{2.585}$

= $2^{2}\times 2^{0.585}$

= 4 × 1.5 [Since, $2^{0.585}=1.5$ ]

= 6

From table 2,

g(x) = $\text{log}_b(x)$

For x = 0.5,

g(0.5) = -1

-1 = $\text{log}_b(0.5)$

b⁻¹ = 0.5

b = 2

For x = 2,

g(2) = 1

1 = $\text{log}_2(2)$

For x = 6,

g(6) = 2.585

2.585 = $\text{log}_2(6)$

2.585 = $\text{log}_2(2\times 3)$

2.585 = $\text{log}_2(2)+\text{log}_2(3)$

2.858 - 1 = $\text{log}_2(3)$

$\text{log}_2(3)=1.585$

For x = 3,

g(3) = $\text{log}_2(3)$

g(3) = 1.585

8 0

3 years ago