Please answer this question! I will try to give brainlist if possible. I beg you, I have asked this question three times!

Evaluate 4/3(4 1/5 + x) ÷ 5/2y when x = 3/2 and y = 15

MissTica

So what u do (-72828282)+(-7382828)=(-73838)+n=(-839288n)

8 0

3 years ago

The nutty professor sells cashews for 7.20 per pound snd brazil nuts flr 4.00 per pound. How much of each type should be used to

lutik1710 [3]

Answer:

<u>19.01 pound</u> of cashews and <u>16.99 pounds</u> of brazil nuts should be used to make a 36 pound mixture that sells for 5.69 per pound.

Step-by-step explanation:

Given:

The nutty professor sells cashews for 7.20 per pound and brazil nuts for 4.00 per pound.

Now, to find each type should be used to make a 36 pound mixture that sells for 5.69 per pound.

Let the quantity of cashews be $x.$

Let the quantity of brazil nuts be $y.$

So, total pound of mixture of nuts:

$x+y=36\\y=36-x\ \ \ .....(1)$

Now, the total price of mixture per pound:

$7.20(x)+4.00(y)=5.69(36)$

$7.2x+4y=204.84$

Substituting the value of $y$ from equation (1) we get:

$7.2x+4(36-x)=204.84$

$7.2x+144-4x=204.84$

$3.2x+144=204.84$

Subtracting both sides by 144 we get:

$3.2x=60.84$

Dividing both sides by 3.2 we get:

$x=19.01.$

The quantity of cashews = 19.01 pound.

Now, substituting the value of $x$ in equation (1) we get:

$y=36-x\\y=36-19.01\\y=16.99.$

The quantity of brazil nuts = 16.99 pounds.

Therefore, 19.01 pound of cashews and 16.99 pounds of brazil nuts should be used to make a 36 pound mixture that sells for 5.69 per pound.

7 0

4 years ago

Can you sleep 7 hours a night how many hours are you awake in a year

horrorfan [7]

Answer:

6,205 hours you are awake in a year

Step-by-step explanation:

5 0

3 years ago

PLS ANSWER DUE

Zinaida [17]

Answer:

Option C

Step-by-step explanation:

We will analyze the figure and note down the properties given in the figure,

1). LK is a diameter so this line (chord) divides the circle into two arcs measuring 180°.

m(arc LJK) = m(arc LK) = 180°

2). m(∠JKL) = 28°

Therefore, by the property inscribed angle and intercepted arcs,

Intercepted arc (JL) = 2 × (Inscribed angle JKL)

m(arc JL) = 2(26°)

= 52°

Now we will use these two points to get the measure of arc JK.

m(arc JK) + m(arc LK) + m(arc JL) = 360°

m(arc JK) + 180° + 52° = 360°

m(arc JK) = 360° - 232°

= 128°

Option C will be the correct option.

4 0

3 years ago

Measurements of the sodium content in samples of two brands of chocolate bar yield the following results (in grams):

Tpy6a [65]

Answer:

98% confidence interval for the difference μX−μY = [ 0.697 , 7.303 ] .

Step-by-step explanation:

We are give the data of Measurements of the sodium content in samples of two brands of chocolate bar (in grams) below;

Brand A : 34.36, 31.26, 37.36, 28.52, 33.14, 32.74, 34.34, 34.33, 29.95

Brand B : 41.08, 38.22, 39.59, 38.82, 36.24, 37.73, 35.03, 39.22, 34.13, 34.33, 34.98, 29.64, 40.60

Also, $\mu_X$ represent the population mean for Brand B and let $\mu_Y$ represent the population mean for Brand A.

Since, we know nothing about the population standard deviation so the pivotal quantity used here for finding confidence interval is;

P.Q. = $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ ~ $t_n__1+n_2-2$

where, $Xbar$ = Sample mean for Brand B data = 36.9

$Ybar$ = Sample mean for Brand A data = 32.9

$n_1$ = Sample size for Brand B data = 13

$n_2$ = Sample size for Brand A data = 9

$s_p = \sqrt{\frac{(n_1-1)s_X^{2}+(n_2-1)s_Y^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(13-1)*10.4+(9-1)*7.1 }{13+9-2} }$ = 3.013

Here, $s^{2}_X$ and $s^{2} _Y$ are sample variance of Brand B and Brand A data respectively.

So, 98% confidence interval for the difference μX−μY is given by;

P(-2.528 < $t_2_0$ < 2.528) = 0.98

P(-2.528 < $\frac{(Xbar -Ybar) -(\mu_X-\mu_Y)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$ < 2.528) = 0.98

P(-2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(Xbar -Ybar) -(\mu_X-\mu_Y)$ < 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

P( (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ < $(\mu_X-\mu_Y)$ < (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ) = 0.98

98% Confidence interval for μX−μY =

[ (Xbar - Ybar) - 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ , (Xbar - Ybar) + 2.528 * $s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2}$ ]

[ $(36.9 - 32.9)-2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ , $(36.9 - 32.9)+2.528*3.013\sqrt{\frac{1}{13} +\frac{1}{9}$ ]

[ 0.697 , 7.303 ]

Therefore, 98% confidence interval for the difference μX−μY is [ 0.697 , 7.303 ] .

4 0

3 years ago