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2 years ago

9

A student bought artichokes for 8.40, cucumbers for 2.88, lettuce for 4.65, tomatoes for 1.15, and broccoli for 2.50. Find a rea

sonable estimate of the total amount the student spent on vegetables by rounding each price to the nearest dollar..

1 answer:

ankoles [38]2 years ago

7 0

Answer:

20

Step-by-step explanation:

8+3+5+1+3=20

rounded up at .50 or higher

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Which of the following gives a product of x^3 - 3x ?

Vilka [71]

Answer:

Option a

Step-by-step explanation:

Lets verify

RHS

$\\ \rm\longmapsto x(x^2-3)$

Use distributive law
a(b-c)=ab-bc

$\\ \rm\longmapsto x(x^2)-3x$

$\\ \rm\longmapsto x^3-3x$

$\\ \rm\longmapsto LHS$

Hence verifed

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2 years ago

Read 2 more answers

Add. Write your answer as a fraction in simplest form.<br> -4/8 + (-23/18)

abruzzese [7]

Answer:

sorry i need more points

Step-by-step explanation:

8 0

2 years ago

-4n+3>23<br> A. -5<br> B. 5<br> C. -9<br> D. -20<br> E. 20

Pani-rosa [81]

It will be 20 after u subtract

6 0

2 years ago

How to convert 1 1/2

seraphim [82]

Answer:

3/2

Step-by-step explanation:

Step 1

Multiply the denominator by the whole number

2 × 1 = 2

Step 2

Add the answer from Step 1 to the numerator

2 + 1 = 3

Step 3

Write answer from Step 2 over the denominator

3/2

3 0

2 years ago

An article in Knee Surgery, Sports Traumatology, Arthroscopy, "Arthroscopic meniscal repair with an absorbable screw: results an

igomit [66]

Answer:

Mean = 1.57

Variance=0.31

Step-by-step explanation:

To calculate the mean and the variance of the number of successful surgeries (X), we first have to enumerate the possible outcomes:

1) Both surgeries are successful (X=2).

$P(e_1)=0.90*0.67=0.603$

2) Left knee unsuccessful and right knee successful (X=1).

$P(e_2)=(1-0.9)*0.67=0.1*0.67=0.067$

3) Right knee unsuccessful and left knee successful (X=1).

$P(e_3)=0.90*(1-0.67)=0.9*0.33=0.297$

4) Both surgeries are unsuccessful (X=0).

$P(e_4)=(1-0.90)*(1-0.67)=0.1*0.33=0.033$

Then, the mean can be calculated as the expected value:

$M=\sum p_iX_i=0.603*2+0.067*1+0.297*1+0.033*0\\\\M=1.206+0.067+0.297+0\\\\M=1.57$

The variance can be calculated as:

$V=\sum p_i(X_i-\bar{X})^2\\\\V=0.603(2-1.57)^2+(0.067+0.297)*(1-1.57)^2+0.033*(0-1.57)^2\\\\V=0.603*0.1849+0.364*0.3249+0.033*2.4649\\\\V=0.1115+0.1183+0.0813\\\\V=0.3111$

3 0

3 years ago