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Afina-wow [57]
3 years ago
12

Between noon and 9pm the temperature dropped 10°F. If the temperature was -4°F at 9pm what was the temperature at noon ?

Mathematics
1 answer:
miv72 [106K]3 years ago
3 0

Answer:

6F

Step-by-step explanation:

Let x be the temperature at noon

dropping means we subtract

The temperature at 9 pm was -4.  That goes on the right hand side of the equals

x-10 = -4

Add 10 to each side

x-10+10 = -4+10

x = 6

The temperature at noon was 6 F

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O is the centre of the circle, EF is a tangent, angle BCE=28 degrees, angle ACD=31 degrees.
Yuki888 [10]

Answer:

a. 28˚

b. 76˚

c. 104˚

d. 56˚

Step-by-step explanation

a. because angle between tangent and chord is equal to the angle(s) in alternate segment.

b. because angles in triangles add up to 180˚, 180-28=152 and because isosceles triangle, 152/2=76˚

c. because angles in triangles add up to 180˚ and opposite angles in a cyclic quadrilateral add up to 180˚, 31+76=107, 180-107=73, 73-28=45, angles in triangle so 180-(31+45)=104˚

d. 28*2=56˚ because angles at circumference are half angles at centre

4 0
3 years ago
The sum of three consecutive counting numbers is 15 more than the greatest
Alex777 [14]

Answer:

9

Step-by-step explanation:

x+x+1+x+2 = 15+x+2

3x+3=x+17

3x=x+14

2x=14

x=7

7+8+9=24

9+15 = 24

7 0
3 years ago
Two machines are used for filling plastic bottles to a net volume of 16.0 ounces. A member of the quality engineering staff susp
aleksandrvk [35]

Answer:

Step-by-step explanation:

Hello!

The objective is to test if the two machines are filling the bottles with a net volume of 16.0 ounces or if at least one of them is different.

The parameter of interest is μ₁ - μ₂, if both machines are filling the same net volume this difference will be zero, if not it will be different.

You have one sample of ten bottles filled by each machine and the net volume of the bottles are measured, determining two variables of interest:

Sample 1 - Machine 1

X₁: Net volume of a plastic bottle filled by the machine 1.

n₁= 10

X[bar]₁= 16.02

S₁= 0.03

Sample 2 - Machine 2

X₂: Net volume of a plastic bottle filled by the machine 2

n₂= 10

X[bar]₂= 16.01

S₂= 0.03

With p-values 0.4209 (for X₁) and 0.6174 (for X₂) for the normality test and using α: 0.05, we can say that both variables of interest have a normal distribution:

X₁~N(μ₁;δ₁²)

X₂~N(μ₂;δ₂²)

Both population variances are unknown you have to conduct a homogeneity of variances test to see if they are equal (if they are you can conduct a pooled t-test) or different (if the variances are different you have to use the Welche's t-test)

H₀: δ₁²/δ₂²=1

H₁: δ₁²/δ₂²≠1

α: 0.05

F= \frac{S^2_1}{S^2_2} * \frac{Sigma^2_1}{Sigma^2_2} ~~F_{n_1-1;n_2-1}

Using a statistic software I've calculated the test

F_{H_0}= 1.41

p-value 0.6168

The p-value is greater than the significance level, so the decision is to not reject the null hypothesis then you can conclude that both population variances for the net volume filled in the plastic bottles by machines 1 and 2 are equal.

To study the difference between the population means you can use

t= \frac{(X[bar]_1-X[bar]_2)-(Mu_1-Mu_2}{Sa\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }  ~~t_{n_1+n_2-2}

The hypotheses of interest are:

H₀: μ₁ - μ₂ = 0

H₁: μ₁ - μ₂ ≠ 0

α: 0.05

Sa= \sqrt{\frac{(n_1-1)S^2_1+(N_2-1)S^2_2}{n_1+n_2-2}} = \sqrt{\frac{9*9.2*10^{-4}+9*6.5*10^{-4}}{10+10-2} } = 0.028= 0.03

t_{H_0}= \frac{(16.02-16.01)-0}{0.03*\sqrt{\frac{1}{10} +\frac{1}{10} } } = 0.149= 0.15

The p-value for this test is: 0.882433

The p-value is greater than the significance level, the decision is to not reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to reject the null hypothesis. You can conclude that the population average of the net volume of the plastic bottles filled by machine one and by machine 2 are equal.

I hope this helps!

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Look at the equation below. Identify which equation has no solution
IgorC [24]

Answer:

C

Step-by-step explanation:

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4 0
2 years ago
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