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2 years ago

The following are the temperatures in °C for the first 10 days in January:

Mathematics

1 answer:

kondor19780726 [428]2 years ago

4 0

The highest temperature is 8.3, in which we call the max.

The lowest temperature is -8.2, which is the min.

Subtract those values to get the range.

Range = max - min = 8.3 - (-8.2) = 8.3 + 8.2 = 16.5

<h3>Answer: 16.5 degrees Celsius</h3>

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An oblique hexagonal prism has a base area of 42

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The base area of the oblique hexagonal prism = 42 cm²

the height of the prism = 4 cm

Volume of the prism = Base area × height = 42 cm² × 4 cm = 168 cm³

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What is the volume (in cubic units) of a sphere with a radius of 18 units? Assume that π = 3.14 and round your answer to the nea

Nataly [62]

Answer:

The volume of sphere is $24416.64\ cm^3$ .

Step-by-step explanation:

We have,

Radius of a sphere is 18 units.

It is required to find the volume of sphere.

The formula of the volume of sphere in terms of radius is given by :

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3 years ago

A researcher tests five individuals who have seen paid political ads about a particular issue. These individuals take a multiple

devlian [24]

Answer:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$

The degrees of freedom are given by:

$df=n-1=5-1=4$

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

Step-by-step explanation:

Information given

48, 41, 40, 51, and 50

The sample mean and deviation can be calculated with these formulas:

$\bar X= \frac{\sum_{i=1}^n X_i}{n}$

$s =\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=46$ represent the mean height for the sample

$s=5.148$ represent the sample standard deviation

$n=5$ sample size

$\mu_o =40$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test

Hypothesis to test

We want to test if the true mean for this case is equal to 40, the system of hypothesis would be:

Null hypothesis: $\mu = 40$

Alternative hypothesis: $\mu \neq 40$

The statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

Replacing we got:

$t=\frac{46-40}{\frac{5.148}{\sqrt{5}}}=2.606$

The degrees of freedom are given by:

$df=n-1=5-1=4$

The p value wuld be given by:

$p_v =2*P(t_{(4)}>2.606)=0.060$

For this case the p value is higher than the significance level so then we can conclude that the true mean is not significantly different from 40

The distribution with the critical values are in the figure attached

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3 years ago