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3 years ago

How many solutions does the following equation have?

Mathematics

1 answer:

Tema [17]3 years ago

6 0

Answer:

no solution

Step-by-step explanation:

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a delivery company charges an extra fee for a package that weighs more than 100 pounds write an inequality that shows the weight

Alona [7]

Answer:

$w\leq 100\ pounds$

Step-by-step explanation:

Let

w------> the weight in pounds

we know that

A delivery company charges an extra fee for a package that weighs more than 100 pounds

For

$w> 100\ pounds$ -----> the company charges an extra fee

therefore

For

$w\leq 100\ pounds$ ----> the company not charge an extra fee

8 0

3 years ago

Help me out please thanks

SVEN [57.7K]

Answer:

1.5 is the correct answer

Step-by-step explanation:

The time it four hours. In those 4 hours the temperature changer by 6 degrees. So 6 divided by 4 is 1.5.

3 0

3 years ago

Read 2 more answers

Such is biger 4/8 or 4/6

MArishka [77]

If you are asking which is bigger then it is 4/6

4 0

4 years ago

If x = a sin α, cos β, y = b sin α.sin β and z = c cos α then (x²/a²) + (y²/b²) + (z²/c²) = ?

Oduvanchick [21]

$\large\underline{\sf{Solution-}}$

<u>Given:</u>

$\rm \longmapsto x = a \sin \alpha \cos \beta$

$\rm \longmapsto y = b \sin \alpha \sin \beta$

$\rm \longmapsto z = c\cos \alpha$

Therefore:

$\rm \longmapsto \dfrac{x}{a} = \sin \alpha \cos \beta$

$\rm \longmapsto \dfrac{y}{b} = \sin \alpha \sin \beta$

$\rm \longmapsto \dfrac{z}{c} = \cos \alpha$

Now:

$\rm = \dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} }$

$\rm = { \sin}^{2} \alpha \cos^{2} \beta + { \sin}^{2} \alpha \sin^{2} \beta + { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha (\cos^{2} \beta + \sin^{2} \beta )+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha \cdot1+ { \cos}^{2} \alpha$

$\rm = { \sin}^{2} \alpha + { \cos}^{2} \alpha$

$\rm = 1$

<u>Therefore:</u>

$\rm \longmapsto\dfrac{ {x}^{2} }{ {a}^{2}} + \dfrac{ {y}^{2} }{ {b}^{2} } + \dfrac{ {z}^{2} }{ {c}^{2} } = 1$

5 0

3 years ago

In a recent year, Washington State public school students taking a mathematics assessment test had a mean score of 276.1 and a s

Oksi-84 [34.3K]

Answer:

a) $\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

b) From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

c) $P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

$P(Z\geq2.070)=1-P(Z$

Step-by-step explanation:

Let X the random variable the represent the scores for the test analyzed. We know that:

$\mu=E(X) = 276.1 , \sigma=Sd(X) = 34.4$

And we select a sample size of 64.

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Part a

For this case the mean and standard error for the sample mean would be given by:

$\mu_{\bar x} =\mu = 276.1$

$\sigma_{\bar x} =\frac{\sigma}{\sqrt{n}}=\frac{34.4}{\sqrt{64}}=4.3$

Part b

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu=276.1, \frac{\sigma}{\sqrt{n}}=4.3)$

Part c

For this case we want this probability:

$P(\bar X \geq 285)$

And we can use the z score defined as:

$z=\frac{\bar x -\mu}{\sigma_{\bar x}}$

And using this we got:

$P(\bar X \geq 285)=P(Z\geq \frac{285-276.1}{4.3}=2.070)$

And using a calculator, excel or the normal standard table we have that:

$P(Z\geq2.070)=1-P(Z$

8 0

3 years ago