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2 years ago

(tap the picture in order to see the hole thing) guys i really need help plz its so annoying that i cant answer it

Mathematics

1 answer:

vlabodo [156]2 years ago

6 0

David's total income is ($8/hr)x + ($11/hr)y.

This problem requires us to write and solve an inequality:

8x + 11(15)<$200

Solving for x: 8x < $200-$165, or 8x < $35. Strictly speaking, x = 4.375 hrs, but if we assume that David must work an integer # of hours, then x=5 full hours.

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An ice cream store sells 3drinks, in 4sizes, and 8 flavors. In how many ways can a customer order a drink?

Sedbober [7]

You would take 3 x 4 which equals 12. And then you would take 12 x8 which would equal 96

7 0

3 years ago

Please help it’s a test and I can’t get logged out

HACTEHA [7]

Answer:

the anwer is B ( i mean second option)

And you can try it

you will find ;

$y = \frac{x}{3} - 1$

HAVE A NİCE DAY

Step-by-step explanation:

GREETİNGS FROM TURKEY ツ

5 0

3 years ago

Find the limit of the function by using direct substitution.

musickatia [10]

Lim x->1 (x^2+8x-2)

substitute x =1
= 1^2 +8(1)-2
= 1 +8 -2
= 7

So the limit = 7

5 0

3 years ago

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 i

stich3 [128]

Complete question :

The annual rainfall in a certain region is approximately normally distributed with mean 41.4 inches and standard deviation 5.7 inches. Round answers to the nearest tenth of a percent. a) What percentage of years will have an annual rainfall of less than 43 inches? b) What percentage of years will have an annual rainfall of more than 39 inches? c) What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

Answer:

0.61053

0.66314

0.26647

Step-by-step explanation:

Given that :

Mean (m) = 41.4 inches

Standard deviation (s) = 5.7 inches

a) What percentage of years will have an annual rainfall of less than 43 inches?

P(x < 43)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (43 - 41.4) / 5.7 = 0.2807017

p(Z < 0.2807) = 0.61053 ( Z probability calculator)

b) What percentage of years will have an annual rainfall of more than 39 inches? c)

P(x > 39)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (39 - 41.4) / 5.7 = −0.421052

p(Z > −0.421052) = 0.66314 ( Z probability calculator)

What percentage of years will have an annual rainfall of between 38 inches and 42 inches?

P(x < 38)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (38 - 41.4) / 5.7 = −0.596491

p(Z < −0.596491) = 0.27545 ( Z probability calculator)

P(x < 42)

USing the relation to obtain the standardized score (Z) :

Z = (x - m) / s

Z = (42 - 41.4) / 5.7 = 0.1052631

p(Z < 0.1052631) = 0.54192 ( Z probability calculator)

0.54192 - 0.27545 = 0.26647

6 0

2 years ago

A rectangle has a length that is 5 meters greater than the width. If w represents the width, write an expression, in terms of w,

ahrayia [7]

Answer:

Area of rectangle is $w^2+5w$

Perimeter of Rectangle is $4w+20$ .

Step-by-step explanation:

Given:

Let the width of the rectangle be 'w'.

Also Given:

A rectangle has a length that is 5 meters greater than the width.

Length of rectangle = $w+5\ meters$

We need to write expression for Area of rectangle and Perimeter of rectangle.

Solution:

Now we know that;

Perimeter of rectangle is equal to twice the sum of the length and width.

framing in equation form we get;

Perimeter of rectangle = $2(w+5+w)=2(2w+5) =4w+20$

Also We know that;

Area of rectangle is length times width.

framing in equation form we get;

Area of rectangle= $w(w+5) = w^2+5w$

Hence Area of rectangle is $w^2+5w$ and Perimeter of Rectangle is $4w+20$ .

6 0

2 years ago