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3 years ago

What is the domain of the function shown is the table

Mathematics

1 answer:

Hoochie [10]3 years ago

7 0

B.(2,4,6,8) is the answer

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Attachment Below Thanks

Viktor [21]

Answer:

Desmond's spinner.

8 0

3 years ago

Read 2 more answers

A gas at standard temperature and pressure was measured as having a mass of 1485 grams (g) in a volume of 750 liters (L). Which

tangare [24]

Answer:

Here we will use the relation:

Density = mass/volume.

We know that:

Mass = 1485g

Volume = 750L

Density = 1485g/750L = 1.98 g/L

We do not have the options in order to select the correct one, but knowing this density, we can suppose the possible gases.

Two "common" gases with densities similar to this one are:

Carbon Dioxide, with d = 1.977 g/L

Dinitrogen Monoxide, with d = 1.977 g/L

In both cases, if we round up we will get the same density that we calculated at the beginning, so either of these can be the correct option.

6 0

3 years ago

Help ill give brainlest

exis [7]

answer: hm i dnt do this at my school but, idrk

just switch them around sorry

Step-by-step explanation:

3 0

2 years ago

The local public transport authority is conducting research into commuter behaviour. A previous census indicated that the mean c

kondor19780726 [428]

Answer:

$H_0:\mu = 45$ and $H_a:\mu < 45$

$T = 34.88$

The null hypothesis is not rejected

Step-by-step explanation:

Given

Random sample of 50 commuter times

Solving (a): The null and alternate hypothesis

From the question, a previous census shows that:

$\mu =45$ and $\sigma= 9.1$

This is the null hypothesis i.e.

$H_0:\mu = 45$

Reading further, a new test was to show whether the average has decreased.

This is the alternate hypothesis, and it is represented as:

$H_a:\mu < 45$

Solving (b) and (c): The test statistic (T).

This is calculated using:

$T = \frac{\mu}{\sigma/\sqrt n}$

In this case:

$n = 50$ $\mu =45$ and $\sigma= 9.1$

So:

$T = \frac{45}{9.1/\sqrt{50}}$

$T = \frac{45}{9.1/7.07}$

$T = \frac{45}{1.29}$

$T = 34.88$

Solving (c): Accept or reject null hypothesis

$\alpha = 0.05$

First calculate the degrees of freedom (df)

$df = n - 1 = 50 - 1$

$d f = 49$

From the t distribution table,

At: $d f = 49$ , $\alpha = 0.05$ and $T = 34.88$

the t score is:

$p = 1.676$

$p > 0.05$

<em>So, the null hypothesis is not rejected</em>

7 0

3 years ago

WILL GIVE Brainelst

Alexeev081 [22]

Step-by-step explanation:

5(m+3)=5m+15

5m+15=5m+15

m=1

6 0

3 years ago