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3 years ago

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Rob can mop a warehouse in 10 hours. Maria can mop the same warehouse in 11 hours. Find how long it would take them if they work

ed together.

2 answers:

Natali [406]3 years ago

4 0

The answer would be 21 good luck hope you get it correct

Wittaler [7]3 years ago

3 0

Answer:

It would take them $\frac{110}{21}$ or approximately 5.24 hours together

Step-by-step explanation:

Use the algebra work formula, $\frac{1}{tb}$ = $\frac{1}{t1}$ + $\frac{1}{t2}$ , where tb is the time together, t1 is the time of one person individually, and t2 is the time of the other person.

Plug in 10 and 11:

$\frac{1}{tb}$ = $\frac{1}{t1}$ + $\frac{1}{t2}$

$\frac{1}{tb}$ = $\frac{1}{10}$ + $\frac{1}{11}$

Find the common denominator, then add the fractions together:

$\frac{1}{tb}$ = $\frac{11}{110}$ + $\frac{10}{110}$

$\frac{1}{tb}$ = $\frac{21}{110}$

Cross multiply and solve for tb:

110 = 21tb

5.24 = tb

So, it would take them $\frac{110}{21}$ or approximately 5.24 hours together

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$\sf Slope=\dfrac{y_2-y_1}{x_2-x_1}$

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3 years ago

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the e

I am Lyosha [343]

Answer:

(a) The proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b) The 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

Step-by-step explanation:

Let <em>X</em> = number of students who read above the eighth grade level.

(a)

A sample of <em>n</em> = 269 students are selected. Of these 269 students, <em>X</em> = 224 students who can read above the eighth grade level.

Compute the proportion of students who can read above the eighth grade level as follows:

$\hat p=\frac{X}{n}=\frac{224}{269}=0.8327$

The proportion of students who can read above the eighth grade level is 0.8327.

Compute the proportion of tenth graders reading at or below the eighth grade level as follows:

$1-\hat p=1-0.8327$

$=0.1673$

Thus, the proportion of tenth graders reading at or below the eighth grade level is 0.1673.

(b)

the information provided is:

<em>n</em> = 709

<em>X</em> = 546

Compute the sample proportion of tenth graders reading at or below the eighth grade level as follows:

$\hat q=1-\hat p$

$=1-\frac{X}{n}$

$=1-\frac{546}{709}$

$=0.2299\\\approx 0.229$

The critical value of <em>z</em> for 95% confidence interval is:

$z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96$

Compute the 95% confidence interval for the population proportion as follows:

$CI=\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p(1-\hat p)}{n}}$

$=0.229\pm 1.96\times \sqrt{\frac{0.229(1-0.229)}{709}}\\=0.229\pm 0.03136\\=(0.19764, 0.26036)\\\approx (0.198, 0.260)$

Thus, the 95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level is (0.198, 0.260).

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3 years ago

Consider the following pair of points.

dedylja [7]

Answer:

The answer is

<h2> $\sqrt{265} \: \: or \: \: 16.28 \: \: \: \: units$ </h2>

Step-by-step explanation:

The distance between two points can be found by using the formula

<h3> $d = \sqrt{ ({x1 - x2})^{2} + ({y1 - y2})^{2} } \\$ </h3>

where

(x1 , y1) and (x2 , y2) are the points

From the question the points are

(9,-8) and (-7,-5)

The distance between them is

$d = \sqrt{ ({9 + 7})^{2} + ({ - 8 + 5})^{2} } \\ = \sqrt{ {16}^{2} + ( { - 3})^{2} } \\ = \sqrt{256 + 9} \: \: \: \: \: \: \: \: \: \\ = \sqrt{265} \: \: \: \: \: \: \\ = 16.27882$

We have the final answer as

$\sqrt{265} \: \: or \: \: 16.28 \: \: \: \: units$

Hope this helps you

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Find the zeros of polynomial function and solve polynomials equations.<br> f(x)=24x^3-64x^2-21x+56

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Since this polynomial has 4 terms, factoring by grouping should be the first thing we try here.

So, we have:

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So, we can use ZPP to find out roots:

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So our three roots are:

$x \in \{\frac{8}{3},\frac{\sqrt{56}}{8},-\frac{\sqrt{56}}{8}\}$

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