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kipiarov [429]
3 years ago
5

Which of the following temperatures is most precise?

Mathematics
1 answer:
Harlamova29_29 [7]3 years ago
8 0
"<span>45.5 degrees" is the temperature among the choices given in the question that is most precise. The correct option among all the options that are given in the question is the fourth option or the last option or option "D". I hope that this is the answer that has actually come to your desired help.</span>
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Henry bought a magazine for $5 and two vitamin waters. Henry spent a total of $11. How much did each vitamin water cost? $
saveliy_v [14]
5+x=11
-5    -5
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x=6
5 0
3 years ago
Read 2 more answers
Please help. I've been stuck on this for 3 days. I just really need help. ​
Marysya12 [62]

Answer:

71.5

Step-by-step explanation:

x+1 so (1+10) x (21+5)= 286

but then 1+1=2 and 2x2= 4 so 286 divided by 4= 71.5

5 0
3 years ago
(b)<br> 2sin(theta) + 1/sin(theta) - 1<br> = 0 for 0 &lt;_ x &lt; 2x
Lady bird [3.3K]

Answer:

2 - 2 cos²x = sin x

2 = sin x + 2 cos² x

0 = -2 + sin x + 2(1 - sin² x)

0 = -2 sin² x + sin x + 2 - 2

0 = 2 sin² x - sin x 0 = sin x (2 sin x - 1)

sin x = 0 v sin x = 0.5

x10° + k.360° x1 = 0°, 360°

x2 = (180° - 0°) + k.360°

x2 = 180°

x3 = 30° + k.360°

x3 = 30°

x4 = (180° -30°) + k.360° x4 = 150°

HP: {x | 0°, 30°, 150°, 180°, 360°)

8 0
2 years ago
The strength of a certain type of rubber is tested by subjecting pieces of the rubber to an abrasion test. For the rubber to be
My name is Ann [436]

Answer:

Step-by-step explanation:

Hello!

You have the hypothesis that the average weight loss for rubber after an abrasion test is less than 3.5 mg. To test this a large sample of pieces of rubber were sampled and subjected to the abrasion test.

With the given information you must test whether the researcher's hypothesis is sustained or not.

The study variable is,

X: Weight loss of rubber cured in a certain way after being subjected to the abrasion test. (mg)

There is no information about the variable distribution, but since it is said that the sample is a "large number" I'll take it as if it is bigger than 30 and apply the Central Limit Theorem to use the approximation of the sample mean to normal. This way I can use the Z-statistic for the test.

Symbolically the statistic hypothesis is:

H₀: μ ≥ 3.5

H₁: μ < 3.5

α: 0.05 (since is not listed, I'll choose one of the most common signification levels)

You have a one-tailed critical region, this means the p-value will also be one-tailed to the left of the distribution (i.e. →-∞)

The formula of the statistic is:

Z= <u> X[bar] - μ </u> ≈ N(0;1)

        δ/√n

To calculate the statistic you have to use the information given.

The sample mean X[bar]= 3.4 mg

Upper bond of 95% CI= 3.45 mg

The basic structure of a CI for the mean is

"estimator" ± "margin of error"

Upper bound is "estimator" + "margin of error"

Using the formula:

Ub= X[bar] + d ⇒ 3.45= 3.4 + d

⇒ d= 3.45 - 3.4 = 0.05

Where d is the margin of error

d= Z_{1-\alpha /2} * (δ/√n)

d= Z_{0.975} * (δ/√n)

d/Z_{0.975}= (δ/√n)

(δ/√n)= 0.05/ 1.96 = 0.0255

(δ/√n) is the denominator in the formula, corresponds to the standard deviation of the distribution.

Now you have all values and can calculate the statistic under the null hypothesis:

Z= <u> 3.4 - 3.5 </u> = -3.92

       0.0255

And the p-value:

P(Z ≤ -3.92) = 0.000044 ⇒ My Z- table goes up to P(Z ≤ -3.00) = 0.001, so using strictly the table I can say that the probability is less than 0.001.

To calculate the exact probability I've used a statistic program.

p-value < 0.001

I hope it helps!

8 0
3 years ago
how to determine whether the equation defines y as a function of x. is it a function yes or no to the problem: x=y^2-6 and probl
kari74 [83]
A\ function\ is\ a\ special\ relationship\ between\ values:\\\\ to\ each\ value\ of\ x\ the\ rule\ assigns\ one\ and\ only\ one\ value\ of\ y.\\------------------------------\\y=x^2+6\ \ \ it\ is\ the\ function\\\\x=0\ \ \ \Rightarrow\ \ \ y=0^2+6=6\\x=-3\ \ \ \Rightarrow\ \ \ y=(-3)^2+6=9+6=15\\x=5\ \ \ \Rightarrow\ \ \ y=5^2+6=25+6=31\\------------------------------\\x=y^2-6\ \ \ it\ is\ not\ the\ function\\

for\ example:\\x=3\ \ \ \ \Rightarrow\ \ \ \ 3=y^2-6\ \ \ \ \Rightarrow\ \ \ y^2=9\ \ \ \ \Rightarrow\ \ \ (y=3\ \ \ or\ \ \ y=-3)\\x=10\ \ \ \Rightarrow\ \ \ 10=y^2-6\ \ \ \Rightarrow\ \ \ y^2=16\ \ \ \Rightarrow\ \ \ (y=4\ \ \ or\ \ \ y=-4)\\x=75\ \ \ \Rightarrow\ \ \ 75=y^2-6\ \ \ \Rightarrow\ \ \ y^2=81\ \ \ \Rightarrow\ \ \ (y=9\ \ \ or\ \ \ y=-9)
3 0
3 years ago
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