Ask question

Ask question

All categories

3 years ago

8

Over the past month, a hairstylist has had 28 female clients and 12 male clients. What is the ratio of male clients to female cl

ients for this hairstylist?
A. 3:7
B. 7:3
C. 7:10
D. 3:10

1 answer:

maw [93]3 years ago

4 0

Answer:its A. 3/7 because it would be 12:28 so you'd simplify it.

Step-by-step explanation:

12/28=3/7

You might be interested in

Write an equation for the horizontal line that contains point E(-3, -1)

ryzh [129]

A horizontal line is represented by y = a number...with that number being the y coordinate of ur point (-3,-1).

Therefore, ur equation is : y = -1

5 0

3 years ago

I need some help with this one. Please give it a go! Thank you for your time!

LekaFEV [45]

Answer:

A) $3x^{2} - x - 4$

B) $-3x^{2} -4x -5$

Step-by-step explanation:

5 0

3 years ago

Which answer describes the calculations that could be used to solve this problem?

Zanzabum

THe answer to ur problem is:

A

8 0

3 years ago

Read 2 more answers

Granny Smith is making strawberry jelly. She fills 6.5 ounce jars to give as gifts. Granny was able to fill 9.5 jars. How many o

ehidna [41]

Granny Smith was able to make 61.75 ounces of strawberry jelly

5 0

3 years ago

The heat index I is a measure of how hot it feels when the relative humidity is H (as a percentage) and the actual air temperatu

PSYCHO15rus [73]

Answer:

a) $I(95,50) = 73.19$ degrees

b) $I_{T}(95,50) = -7.73$

Step-by-step explanation:

An approximate formula for the heat index that is valid for (T ,H) near (90, 40) is:

$I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}$

a) Calculate I at (T ,H) = (95, 50).

$I(95,50) = 45.33 + 0.6845*(95) + 5.758*(50) - 0.00365*(95)^{2} - 0.1565*95*50 + 0.001*50*95^{2} = 73.19$ degrees

(b) Which partial derivative tells us the increase in I per degree increase in T when (T ,H) = (95, 50)? Calculate this partial derivative.

This is the partial derivative of I in function of T, that is $I_{T}(T,H)$ . So

$I(T,H) = 45.33 + 0.6845T + 5.758H - 0.00365T^{2} - 0.1565TH + 0.001HT^{2}$

$I_{T}(T,H) = 0.6845 - 2*0.00365T - 0.1565H + 2*0.001H$

$I_{T}(95,50) = 0.6845 - 2*0.00365*(95) - 0.1565*(50) + 2*0.001(50) = -7.73$

8 0

3 years ago