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3 years ago

Help please I'm not getting this at all. I'll send the whole picture

Mathematics

2 answers:

Vikentia [17]3 years ago

8 0

Answer:

true. hope this helps!!

Hunter-Best [27]3 years ago

6 0

The answer is definitely True

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The graph shows the relationship between the number of hours that Michelle has been driving and the distance that she has left t

ahrayia [7]

ANSWER:yes it is hopefully u get it right comment if it’s wrong

7 0

3 years ago

Read 2 more answers

The quotient of (x4 + 5x3 – 3x – 15) and a polynomial is (x3 – 3). What is the polynomial?

grandymaker [24]

Answer:

g(x) = x+5

Step-by-step explanation:

Given that,

$f(x)=x^4+5x^3-3x-15$

$q(x) = (x^3-3)$

We need to find the polynomial. We know that, Euclid division lemma states that

$f(x)=q(x)\times g(x)+r(x)$

Where

f(x) is dividend

g(x) is the poynomial

q(x) is quotient

r(x) is remainder

So,

$x^4+5x^3-3x-15=(x^3-3)\times g(x) +0\\\\g(x)=\dfrac{x^4+5x^3-3x-15}{(x^3-3)}\\\\g(x)=x+5$

So, the polynomial is (x+5).

3 0

3 years ago

Read 2 more answers

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

3 years ago

At the end of the summer, liam discovers that his radiator antifreeze solution has dropped below the safe level. If the radiator

attashe74 [19]

2 gallons 25% of 4 is 1 so there is about 1 gallon of pure in there no so adding two will make 3 gallons of antifreeze bringing total to 6 gallons 3/6 = .5 boom bam 50%

4 0

4 years ago

Use the intermediate value theorem to determine whether the polynomial function has a real zeros between the given integers.

zavuch27 [327]

$f(1)=-5\cdot1^4+8\cdot1^3+6\cdot1+1=-5+8+6+1=10\\
f(2)=-5\cdot2^4+8\cdot2^3+6\cdot2+1=-80+64+12+1=-3\\\\
f(1)>0\\
f(2)$
It does.

3 0

4 years ago

Help please I'm not getting this at all. I'll send the whole picture ​

Help please I'm not getting this at all. I'll send the whole picture