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3 years ago

It takes Wayne 1/0 of an hour to walk a 1/6 mile park loop. What is Wayne's unit rate in miles per hour?

Mathematics

1 answer:

Aleksandr [31]3 years ago

4 0

1 2/3 miles is the answer.

I think you meant 1/10

6 + 4 = 10 1 mile 4/6 = 1 2/3

Hope I beecome the brainliest

You might be interested in

6. Find the value of x in the triangle below.

Kobotan [32]

Answer:

The value of x is 21.

Step-by-step explanation:

Since this is a right angle triangle, you can apply Pythagoras theorem, side² + side² = hypo² :

(x + 3)² + 10² = (x + 5)²

x² + 6x + 9 + 100 = x² + 10x + 25

x² + 6x + 109 - x² - 10x - 25 = 0

-4x + 84 = 0

-4x = -84

4x = 84

x = 84 ÷ 4

x = 21 units

4 0

2 years ago

I really need help with this question could I please get help?

Novosadov [1.4K]

X is equal to 12 and y is equal to 8

5 0

3 years ago

You deposit $500 into a savings account that earns 3.2% annual interest. How much is in the savings account after 10 years?

Brilliant_brown [7]

A = P(1 +r/n)^nt
P = 500
r = 3.2% = 0.032
n = 1
t = 10

A = 500(1 + 0.032)^10
A = $685.12

7 0

2 years ago

There are 75 milligrams of cholesterol in a 3.5 ounce serving of lobster. How much cholesterol is in 6ounces of lobster

Maru [420]

Hence, the amount of cholesterol in 6 ounces of lobster will be 128.58 milligrams

Step-by-step explanation:

Given

Cholesterol in 3.5 ounces = 75 mg

Cholesterol in 6 ounces =x= ?

This can be worked out by using proportions

It can be written as:

$\frac{75}{3.5}=\frac{x}{6}\\21.43=\frac{x}{6}\\Multiplying\ both\ sides\ by\ 6\\21.43*6=x\\x=128.58\ mg$

Hence, the amount of cholesterol in 6 ounces of lobster will be 128.58 milligrams

Keywords: Proportions, Conversion

Learn more about proportions at:

#LearnwithBrainly

8 0

3 years ago

A certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder. In

lord [1]

Answer:

95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

Step-by-step explanation:

We are given that a certain geneticist is interested in the proportion of males and females in the population who have a minor blood disorder.

A random sample of 1000 males, 250 are found to be afflicted, whereas 275 of 1000 females tested appear to have the disorder.

Firstly, the pivotal quantity for 95% confidence interval for the difference between population proportion is given by;

P.Q. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of males having blood disorder= $\frac{250}{1000}$ = 0.25

$\hat p_2$ = sample proportion of females having blood disorder = $\frac{275}{1000}$ = 0.275

$n_1$ = sample of males = 1000

$n_2$ = sample of females = 1000

$p_1$ = population proportion of males having blood disorder

$p_2$ = population proportion of females having blood disorder

Here for constructing 95% confidence interval we have used Two-sample z proportion statistics.

So, 95% confidence interval for the difference between the population proportions, ( $p_1-p_2$ ) is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5% level

of significance are -1.96 & 1.96}

P(-1.96 < $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < 1.96) = 0.95

P( $-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ${(\hat p_1-\hat p_2)-(p_1-p_2)}$ < $1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

P( $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ < ( $p_1-p_2$ ) < $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ) = 0.95

95% confidence interval for ( $p_1-p_2$ ) =

[ $(\hat p_1-\hat p_2)-1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ , $(\hat p_1-\hat p_2)+1.96 \times {\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+ \frac{\hat p_2(1-\hat p_2)}{n_2}} }$ ]

= [ $(0.25-0.275)-1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ , $(0.25-0.275)+1.96 \times {\sqrt{\frac{0.25(1-0.25)}{1000}+ \frac{0.275(1-0.275)}{1000}} }$ ]

= [-0.064 , 0.014]

Therefore, 95% confidence interval for the difference between the proportions of males and females who have the blood disorder is [-0.064 , 0.014].

8 0

3 years ago