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3 years ago

Help asap please: Solve x-7y=16 for y

Mathematics

2 answers:

Pepsi [2]3 years ago

5 0

Answer:

y = (x-16)/7

Step-by-step explanation:

x - 7y = 16

~Subtract x to both sides

-7y = 16 - x

~Divide -7 to everything

y = (16-x)/7

Best of Luck!

fenix001 [56]3 years ago

3 0

Answer:

y = 1/7 x -16/7

Step-by-step explanation:

x-7y=16

Subtract x from each side

x-x-7y=16-x

-7y = -x+16

Divide each side by -7

-7y/-7 = -x/-7 +16/-7

y = 1/7 x -16/7

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If y=kx, where k is a constant, and y=24 when x=6, what is the value of y when x=5?A. 6B. 15C. 20D. 23

Marizza181 [45]

First, we will find the value of k

We can do this by sybstituting y=24, x=6 in;

y=kx and then solve for k

24= k(6)

divide both-side of the equation by 6

24/6 = k

4 = k

k=4

Then when x = 5, we will substitute x=5 and k=4 in; y=kx and then solve for y

y= (4)(5)

y = 20

7 0

1 year ago

A company manufactures aluminum mailboxes in the shape of a box with a half-cylinder top. The company will make 1863 mailboxes t

forsale [732]

Answer:

3433 m²

Step-by-step explanation:

From the image, we have a rectangular box without cover and half a cylinder on top.

Formula for surface area of rectangular box with top is;

S = 2(lh + wh + lw)

From the image,

l = 0.6 m

w = 0.4 m

h = 0.55 m

Thus;

S = 2((0.6 × 0.55) + (0.4 × 0.55) + (0.6 × 0.4))

S = 1.58 m²

Now, since the top is not included for this figure, then;

Surface area of this rectangular box is;

S1 = 1.58 - (lw) = 1.58 - (0.4 × 0.6) = 1.34 m²

Surface area of a cylinder is;

S = 2πr² + 2πrh

r is radius and in this case = 0.4/2 = 0.2 m

h = 0.6

S = 2π(0.2² + (0.2 × 0.6))

S = 1.005 m²

Since it is half cylinder, then we have;

S2 = 1.005/2

S2 = 0.5025 m²

Total surface area; S_t = S1 + S2

S_t = 1.34 + 0.5025

S_t = 1.8425 m²

This is the surface area of one mail box.

Thus, for 1863 mailboxes, total surface area is;

S = 1863 × 1.8425 = 3432.5775 m²

Approximating to the nearest Sq.m gives;

S = 3433 m²

6 0

3 years ago

Use the area of the rectangle to find the area of the triangle. 10 ft

77julia77 [94]

Answer: 15

Aka the first one

7 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

If f(n) = -6n + 2, find f (5/6)

joja [24]

$f( \frac{5}{6} ) = - 6( \frac{5}{6} ) + 2 \\ f( \frac{5}{6} ) = - \frac{30}{6} + 2 \\ f( \frac{5}{6} ) = - 5 + 2 \\ f( \frac{5}{6} ) = - 3$

7 0

3 years ago

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