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4 years ago

15

sarah bought 3 blouses at one price and 2 blouses prices $3 below the others. The total cost was $91.50. Find the price of the b

louses

1 answer:

jekas [21]4 years ago

4 0

33 30 24 that should be ur answer

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A bag contains 5 white balls and 4 black balls, in how many ways 2 balls be drawn?

Elodia [21]

Answer:

4 ways (WW-BB-WB-BW)

Step-by-step explanation:

3 0

3 years ago

Hi! Please asnwer this for 30 points!! "Jada applies two transformations to a polygon in the coordinate plane. One of the transf

sveticcg [70]

Using translation concepts, we have that:

For the translation, she has to communicate if it is up, down, left or right and the number of units.

For a reflection she must communicate over which line the reflection happened.

<h3>What is a translation?</h3>

A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction either in it’s definition or in it’s domain. Examples are shift left/right or bottom/up, vertical or horizontal stretching or compression, and reflections over the x-axis or the y-axis.

A translation is either shift left/right or bottom/up, hence she has to communicate if it is up, down, left or right and the number of units.

A reflection is over a line, hence she must communicate over which line the reflection happened.

More can be learned about translation concepts at brainly.com/question/28373831

#SPJ1

4 0

2 years ago

A company produces a certain product, and each unit of this product may have 3 different types of defects. Let Di, D2,Ds represe

Stella [2.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

a) 0.88

b) 0.02

c) 0.01

d) 0.99

Step-by-step explanation:

Step one: State the given parameters

$P(D_{1} ) = 0.12$ $P(D_{2} ) = 0.07$

$P(D_{3} ) = 0.05$ $P (D_{1} U D_{2} ) = 0.13$

$P(D_{1}n D_{2}n D_{3}) = 0.01$ $P(D_{1} U D_{3}) = 0.14$

Step 2 : Obtain the probability that a unit does not have a type 1 defect

$P(\frac{}{D_{1} })$ = $1 -P(D_{1} )$

= $1 - 0.12$

= 0.88

Step 3 : Obtain the probability that a unit has both type 2 and 3 defect?

The probability of the unit having both type 2 and type 3 defect is denoted as $P(D_{2} n D_{3} )$

This is calculated as

$P(D_{2}n D_{3}) =P(D_{2} ) + P(D_{3}) - P(D_{2} U D_{3})\\\\ = 0.07 + 0,05 - 0.13$

= 0.02

Therefore P(D_{2} n D_{3} ) = 0.02

Step 4 : Obtain the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect

Let $P(\frac{}{D_{1}} n D_{2} n D_{3} )$ denote the probability that the unit has both a type 2 and type 3 ,but not a type 1 defect.

This can be calculated as follows :

$P(\frac{}{D_{1}} n D_{2} n D_{3} ) = P(D_{2} n D_{3}) - P(D_{1} n D_{2}nD_{3})$

= 0.02 - 0.01

= 0.01

Step 4 : Obtain the probability that a unit has at most two defects

P(at most 2 defects) = 1 - P(all three defects)

= $1- P(D_{1} n D_{2}nD_{3})$

= 1 - 0.01

= 0.99

7 0

3 years ago

A tall tree has a shadow that measures 40 feet. Next to the tree is a man whose shadow measures 4

o-na [289]

<h3>The height of the tree is 55 feet</h3>

<em><u>Solution:</u></em>

Let,

Height of tree = x

Shadow of tree = 40 feet

Height of man = 5.5 feet

shadow of man = 4 feet

Set up a proportion comparing the height of each object to the length of the shadow.

$\frac{\text{ Height of tree}}{\text{ Shadow of tree }} = \frac{\text{ Height of man}}{\text{ shadow of man }}$

Therefore,

$\frac{x}{40} = \frac{5.5}{4}\\\\x = 5.5 \times 10\\\\x = 55$

Thus height of the tree is 55 feet

3 0

3 years ago

Based on the information below, what are the values of x and y of the solution to the system of equations used to create the

ryzh [129]

Answer:

What is the x-value of the solution to the system of equations shown below?

Step-by-step explanation:

y = 3x +5

2y = 4x + 24

6 0

4 years ago