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3 years ago

Which graph represents the linear function y = x 3/4– 4?

Mathematics

1 answer:

ahrayia [7]3 years ago

3 0

Answer:

The graph in the attached figure

Step-by-step explanation:

we have

$y=\frac{3}{4}x-4$

This is a linear equation (the graph is a line)

To identify the graph find out the intercepts

Find out the y-intercept

The y-intercept is the value of y when the value of x is equal to zero

For x=0

$y=\frac{3}{4}(0)-4=-4$

The y-intercept is the point (0,-4)

Find out the x-intercept

The x-intercept is the value of x when the value of y is equal to zero

For y=0

$0=\frac{3}{4}x-4$

$\frac{3}{4}x=4$

$x=16/3=5.33$

The x-intercept is the point (5.33,0)

therefore

The graph in the attached figure

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amm1812

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$\frac{3}{10} \\ \\ or \: in \: alternate \: \: form \\ \\ 0.3$

<h2>refer to the attachment for explanation</h2>

<h2>hope it helps</h2>

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2 years ago

Read 2 more answers

1.What is the true density of water?

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$\huge\color{red}{\boxed{\tt{A}}}\color{orange}{\boxed{\tt{N}}}\color{lime}{\boxed{\tt{S}}}\color{aqua}{\boxed{\tt{W}}}\color{violet}{\boxed{\tt{E}}}\color{pink}{\boxed{\tt{R}}}\color{magenta}$

A common unit of measurement for water's density is gram per milliliter (1 g/ml) or 1 gram per cubic centimeter (1 g/cm3). Actually, the exact density of water is not really 1 g/ml, but rather a bit less (very, very little less), at 0.9998395 g/ml at 4.0° Celsius (39.2° Fahrenheit).
Water density changes with temperature and salinity. Density is measured as mass (g) per unit of volume (cm³). Water is densest at 3.98°C and is least dense at 0°C (freezing point). Water density changes with temperature and salinity.
When water is a liquid, the water molecules are packed relatively close together but can slide past each other and move around freely (as stated earlier, that makes it a liquid). Pure water has a density of 1.000 g/cm3 at 4˚ C. As the temperature increases or decreases from 4˚ C, the density of water decreases.

Step-by-step explanation:

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2 years ago

Read 2 more answers

Help on logarithmic equation

inessss [21]

$\log _{ 2 }{ \left( 3-x \right) } +\log _{ 2 }{ 5 } =2\log _{ 3 }{ 5 } \\ \\ \log _{ 2 }{ \left( 5\left( 3-x \right) \right) } =\log _{ 3 }{ \left( { 5 }^{ 2 } \right) } \\ \\ { 2 }^{ \log _{ 3 }{ 25 } }=5\left( 3-x \right) \\ \\ { 2 }^{ \log _{ 3 }{ 25 } }=15-5x\\ \\ 5x=15-{ 2 }^{ \log _{ 3 }{ 25 } }\\ \\ x=\frac { 15 }{ 5 } -\frac { { 2 }^{ \log _{ 3 }{ 25 } } }{ 5 } \\ \\ x=3-\frac { { 2 }^{ \log _{ 3 }{ 25 } } }{ 5 }$

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3 years ago

A clinical trial tests a method designed to increase the probability of conceiving a girl. In the study 340 babies were born, a

Bas_tet [7]

Answer:

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

b)Yes, the proportion of girls is significantly different from 0.5.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$ , and a confidence interval $1-\alpha$ , we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

Z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$ .

For this problem, we have that:

In the study 340 babies were born, and 289 of them were girls. This means that $n = 340, \pi = \frac{289}{340} = 0.85$

Use the sample data to construct a 99% confidence interval estimate of the percentage of girls born.

So $\alpha = 0.01$ , z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$ , so $z = 2.575$ .

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 - 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.80$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{340}} = 0.85 + 2.575\sqrt{\frac{0.85*0.15}{400}} = 0.90$

The 99% confidence interval estimate of the percentage of girls born is (0.8, 0.9).

Does the method appear to be effective?

b)Yes, the proportion of girls is significantly different from 0.5.

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3 years ago

Which division problem is equivalent to 25?

meriva

They all add up to 2.5 maybe u have to simplify?

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3 years ago