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Ede4ka [16]
3 years ago
10

A beekeeper pours honey into jars. Each large jar holds 18 ounces of honey. Each small jar holds 12 ounces of honey. The beekeep

er fills 32 large jars and 46 small jars.
Which answer is the most reasonable estimate of the total amount of honey in the jars?


800 ounces because 10 × 30 + 10 × 50 = 800

1,000 ounces because 20 × 30 + 10 × 40 = 1,000

1,100 ounces because 20 × 30 + 10 × 50 = 1,100

1,600 ounces because 20 × 30 + 20 × 50 = 1,600
Mathematics
1 answer:
viktelen [127]3 years ago
8 0

Answer:

1,100 ounces because 20 × 30 + 10 × 50 = 1,100

Step-by-step explanation:

You might be interested in
5/6*2/3<br> 9/10*5/18<br> 4/5*3/4<br> 2/3*5/1
zalisa [80]

To multiply fractions, you multiply numerator with numerator and denominator with denominator

In other words: a/b*c/d=(a*c)/(b*d)

So 1st one is (5*2)/(6*3) which is 10/18 or 5/9 simplified

2nd one is (9*5)/(10*18) which is 45/180 or 1/4 simplified

3rd one is (4*3)/(5*4) which is 12/20 or 3/5 simplified

4th one is (2*5)/(3*1) which is 10/3 or 3 1/3 if you want a mixed number

Hope this helped!

5 0
4 years ago
Read 2 more answers
Seventy percent of all vehicles examined at a certain emissions inspection station pass the inspection. Assuming that successive
NeX [460]

Answer:

(a) 0.343

(b) 0.657

(c) 0.189

(d) 0.216

(e) 0.353

Step-by-step explanation:

Let P(a vehicle passing the test) = p

                        p = \frac{70}{100} = 0.7  

Let P(a vehicle not passing the test) = q

                         q = 1 - p

                         q = 1 - 0.7 = 0.3

(a) P(all of the next three vehicles inspected pass) = P(ppp)

                           = 0.7 × 0.7 × 0.7

                           = 0.343

(b) P(at least one of the next three inspected fails) = P(qpp or qqp or pqp or pqq or ppq or qpq or qqq)

      = (0.3 × 0.7 × 0.7) + (0.3 × 0.3 × 0.7) + (0.7 × 0.3 × 0.7) + (0.7 × 0.3 × 0.3) + (0.7 × 0.7 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.3)

      = 0.147 + 0.063 + 0.147 + 0.063 + 0.147 + 0.063 + 0.027

      = 0.657

(c) P(exactly one of the next three inspected passes) = P(pqq or qpq or qqp)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

                 = 0.063 + 0.063 + 0.063

                 = 0.189

(d) P(at most one of the next three vehicles inspected passes) = P(pqq or qpq or qqp or qqq)

                 =  (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7) + (0.3 × 0.3 × 0.3)

                 = 0.063 + 0.063 + 0.063 + 0.027

                 = 0.216

(e) Given that at least one of the next 3 vehicles passes inspection, what is the probability that all 3 pass (a conditional probability)?

P(at least one of the next three vehicles inspected passes) = P(ppp or ppq or pqp or qpp or pqq or qpq or qqp)

=  (0.7 × 0.7 × 0.7) + (0.7 × 0.7 × 0.3) + (0.7 × 0.3 × 0.7) + (0.3 × 0.7 × 0.7) + (0.7 × 0.3 × 0.3) + (0.3 × 0.7 × 0.3) + (0.3 × 0.3 × 0.7)

= 0.343 + 0.147 + 0.147 + 0.147 + 0.063 + 0.063 + 0.063

                  = 0.973  

With the condition that at least one of the next 3 vehicles passes inspection, the probability that all 3 pass is,

                         = \frac{P(all\ of\ the\ next\ three\ vehicles\ inspected\ pass)}{P(at\ least\ one\ of\ the\ next\ three\ vehicles\ inspected\ passes)}

                         = \frac{0.343}{0.973}

                         = 0.353

3 0
4 years ago
Read 2 more answers
Is the relationship shown by the data linear? x -9, -5, -1, 3 y -2,-7,-12, -17
Temka [501]
                x               y
            ------           -----
              -9              -2
             -5               -7
             -1              -12
              3               -17

ur x's all have a difference of 4, and ur y's all have a difference of -5...so yes, this is linear with a slope of (-7+2) / (-5 + 9) = -5/4
5 0
3 years ago
A positive integer is twice another.the sum of the reciprocal of the two positive integer is 3/14. Find the integers
icang [17]

Answer:

\huge\boxed{14\ \text{and}\ 7}

Step-by-step explanation:

n,\ m-\text{positive integer}\\\\n=2m-\text{a positive integer is twice another}\\\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}-\text{the sum of the reciprocal of the two positive integer is }\ \dfrac{3}{14}\\\\\text{We have the system of equations:}\\\\\left\{\begin{array}{ccc}n=2m&(1)\\\dfrac{1}{n}+\dfrac{1}{m}=\dfrac{3}{14}&(2)\end{array}\right

\text{Substitute (1) to (2):}\\\\\dfrac{1}{2m}+\dfrac{1}{m}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{1\cdot2}{m\cdot2}=\dfrac{3}{14}\\\\\dfrac{1}{2m}+\dfrac{2}{2m}=\dfrac{3}{14}\\\\\dfrac{1+2}{2m}=\dfrac{3}{14}\\\\\dfrac{3}{2m}=\dfrac{3}{14}\Rightarrow2m=14\qquad\text{divide both sides by 2}\\\\\dfrac{2m}{2}=\dfrac{14}{2}\\\\\boxed{m=7}

\text{Substitute it to (1):}\\\\n=2\cdot7\\\\\boxed{n=14}

8 0
3 years ago
A set of equations is shown above. Which method eliminates one of the variables? A) Multiply equation A by -1/3 and add the resu
Paraphin [41]

Answer:


Step-by-step explanation:

Multiplying Equation A by (1/3) and adding the result to Equation B will do the trick.  Let's actually solve the problem!

Equation A:  (5/3)x + 3y = 12

Equation B:    4x     -  3y = 8

                    ---------------------------

                     (5/3 + 12/3)x = 15        Note how this has eliminated the variable

                            (17/3)x = 15           y.

                                   x = (3/17)(15)

8 0
3 years ago
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