Please help me! 17 / 13

What fraction of 150 gallons is 189 l? Give your answer in simplest form. 1 gallon = 4.5 l.

lys-0071 [83]

Answer:

$\frac{25}{7}$

Step-by-step explanation:

Given:

1 gallon = 4.5 liters

Question asked:

What fraction of 150 gallons is 189 liters ?

Solution:

First of all we will convert gallons to liters:-

1 gallon = 4.5 liters

150 gallons = 4.5 $\times$ 150 = 675 liters

Now, to know the fraction :-

675 gallons $\div$ 189 liters

$=\frac{675}{189} \\ \\ Dividng\ both\ numerator\ and\ denominator\ by\ 9\\ \\ =\frac{75}{21} \\ \\ Dividng\ both\ numerator\ and\ denominator\ by\ 3\\ \\ =\frac{25}{7}$

Therefore, 150 gallons (675 liters) is $\frac{25}{7}$ of 189 liters.

5 0

3 years ago

Read 2 more answers

If angle abc is reflected across the y-axis, what are the coordinates of A”?

larisa86 [58]

Answer:

The coordinates of A'' are (-2 , -5) ⇒ answer A

Step-by-step explanation:

* Lets revise some transformation

- If point (x , y) reflected across the x-axis

∴ Its image is (x , -y)

- If point (x , y) reflected across the y-axis

∴ Its image is (-x , y)

- If point (x , y) reflected across the line y = x

∴ Its image is (y , x)

- If point (x , y) reflected across the line y = -x

∴ Its image is (-y , -x)

* Now lets solve the problem

- The vertices of triangle ABC are:

A is (2 , -5) , B is (1 , -3) , C is (5 , -3)

∵ The triangle is reflected across the y-axis

∴ The x- coordinates of the three point are changed to opposite sign

∵ A is (2 , -5)

∴ its image A" is (-2 , -5)

* The coordinates of A'' are (-2 , -5)

8 0

4 years ago

Land's Bend sells a wide variety of outdoor equipment and clothing. The company sells both through mail order and via the intern

Naya [18.7K]

Answer:

80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

Step-by-step explanation:

We are given that a random sample of 7 sales receipts for mail-order sales results in a mean sale amount of $81.70 with a standard deviation of $18.75.

A random sample of 11 sales receipts for internet sales results in a mean sale amount of $74.60 with a standard deviation of $28.25.

Firstly, the Pivotal quantity for 80% confidence interval for the difference between population means is given by;

P.Q. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean sales receipts for mail-order sales = $81.70

$\bar X_2$ = sample mean sales receipts for internet sales = $74.60

$s_1$ = sample standard deviation for mail-order sales = $18.75

$s_2$ = sample standard deviation for internet sales = $28.25

$n_1$ = size of sales receipts for mail-order sales = 7

$n_2$ = size of sales receipts for internet sales = 11

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(7-1)\times 18.75^{2} +(11-1)\times 28.25^{2} }{7+11-2} }$ = 25.11

Here for constructing 80% confidence interval we have used Two-sample t test statistics as we don't know about population standard deviations.

So, 80% confidence interval for the difference between population means, ( $\mu_1-\mu_2$ ) is ;

P(-1.337 < $t_1_6$ < 1.337) = 0.80 {As the critical value of t at 16 degree

of freedom are -1.337 & 1.337 with P = 10%}

P(-1.337 < $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 1.337) = 0.80

P( $-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ${(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}$ < $1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

P( $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < ( $\mu_1-\mu_2$ ) < $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ) = 0.80

80% confidence interval for ( $\mu_1-\mu_2$ ) =

[ $(\bar X_1-\bar X_2)-1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ , $(\bar X_1-\bar X_2)+1.337 \times {s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ]

= [ $(81.70-74.60)-1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ , $(81.70-74.60)+1.337 \times {25.11 \times \sqrt{\frac{1}{7} +\frac{1}{11} } }$ ]

= [-9.132 , 23.332]

Therefore, 80% confidence interval for the true mean difference between the mean amount of mail-order purchases and the mean amount of internet purchases ( $\mu_1-\mu_2$ ) is [-9.132 , 23.332].

4 0

3 years ago

Use the graph below to answer the following question:

Elza [17]

Using the quotient, [f(-4)-f(1)]/(-4-1)=(4--1)/(-4-1)=5/-5=-1 is the average rate of change over [-4,1]

5 0

3 years ago

Find r(t) if r'(t)=ti+e^tj+te^tk and r(0)=i+j+k

weeeeeb [17]

R(t) = integral of r'(t) = integral of ti + e^tj + te^tk = 1/2t^2i + e^tj + (te^t - e^t)k + c
r(0) = j - k + c = i + j + k
c = i + 2k
Therefore, r(t) = (1/2t^2 + 1)i + e^tj + (te^t - e^t + 2)k

6 0

4 years ago

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