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3 years ago

Round 23 to the nearest ten

Mathematics

2 answers:

svetlana [45]3 years ago

7 0

Answer:

23 is between 20 and 30, but it's closer to 20. So 23 is rounded down to 20.

Vladimir [108]3 years ago

7 0

Answer:

Step-by-step explanation:

tens ones

2 3

We look at the ones place when rounding to the tens

3 is less than 5 so we leave the tens place alone ( 5 or higher round the tens place up)

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A pen company averages 1.2 defective pens per carton produced (200 pens). The number of defects per carton is Poisson distribute

nlexa [21]

Answer:

a. P(x = 0 | λ = 1.2) = 0.301

b. P(x ≥ 8 | λ = 1.2) = 0.000

c. P(x > 5 | λ = 1.2) = 0.002

Step-by-step explanation:

If the number of defects per carton is Poisson distributed, with parameter 1.2 pens/carton, we can model the probability of k defects as:

$P(k)=\frac{\lambda^{k}e^{-\lambda}}{k!}= \frac{1.2^{k}\cdot e^{-1.2}}{k!}$

a. What is the probability of selecting a carton and finding no defective pens?

This happens for k=0, so the probability is:

$P(0)=\frac{1.2^{0}\cdot e^{-1.2}}{0!}=e^{-1.2}=0.301$

b. What is the probability of finding eight or more defective pens in a carton?

This can be calculated as one minus the probablity of having 7 or less defective pens.

$P(k\geq8)=1-P(k$

$P(0)=1.2^{0} \cdot e^{-1.2}/0!=1*0.3012/1=0.301\\\\P(1)=1.2^{1} \cdot e^{-1.2}/1!=1*0.3012/1=0.361\\\\P(2)=1.2^{2} \cdot e^{-1.2}/2!=1*0.3012/2=0.217\\\\P(3)=1.2^{3} \cdot e^{-1.2}/3!=2*0.3012/6=0.087\\\\P(4)=1.2^{4} \cdot e^{-1.2}/4!=2*0.3012/24=0.026\\\\P(5)=1.2^{5} \cdot e^{-1.2}/5!=2*0.3012/120=0.006\\\\P(6)=1.2^{6} \cdot e^{-1.2}/6!=3*0.3012/720=0.001\\\\P(7)=1.2^{7} \cdot e^{-1.2}/7!=4*0.3012/5040=0\\\\$

$P(k$

c. Suppose a purchaser of these pens will quit buying from the company if a carton contains more than five defective pens. What is the probability that a carton contains more than five defective pens?

We can calculate this as we did the previous question, but for k=5.

$P(k>5)=1-P(k\leq5)=1-\sum_{k=0}^5P(k)\\\\P(k>5)=1-(0.301+0.361+0.217+0.087+0.026+0.006)\\\\P(k>5)=1-0.998=0.002$

5 0

3 years ago

At which root does the graph of f(x) = (x + 4)(x + 7,5) cross the x-axis? -7 -4 4 7

Alona [7]

Answer:

Answer: -7

Step-by-step explanation:

5 0

2 years ago

It's about ratios I'm just kinda stuck

Blababa [14]

Answer:

Step-by-step explanation:

These answers are in order btw!

1=2

2=4

4=6

8=10

Table 2:

1=3

2=6

3=9

4=12

5=15

I’ll finish table 3 and put it in the commenta since this one will take me longer

8 0

3 years ago

Read 2 more answers

If you were present 120 days out of 124 days what percent were you present

emmainna [20.7K]

Answer:

~96.6

Step-by-step explanation:

120/124 is 30/31 or 60/62 or 96.6/99.82

5 0

3 years ago

The U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542. Suppos

xenn [34]

Answer:

(a) P(X > $57,000) = 0.0643

(b) P(X < $46,000) = 0.1423

(c) P(X > $40,000) = 0.0066

(d) P($45,000 < X < $54,000) = 0.6959

Step-by-step explanation:

We are given that U.S. Bureau of Economic Statistics reports that the average annual salary in the metropolitan Boston area is $50,542.

Suppose annual salaries in the metropolitan Boston area are normally distributed with a standard deviation of $4,246.

Let X = annual salaries in the metropolitan Boston area

SO, X ~ Normal( $\mu=$50,542,\sigma^{2} = $4,246^{2}$ )

The z-score probability distribution for normal distribution is given by;

Z = $\frac{X-\mu}{\sigma }$ ~ N(0,1)

where, $\mu$ = average annual salary in the Boston area = $50,542

$\sigma$ = standard deviation = $4,246

(a) Probability that the worker’s annual salary is more than $57,000 is given by = P(X > $57,000)

P(X > $57,000) = P( $\frac{X-\mu}{\sigma }$ > $\frac{57,000-50,542}{4,246 }$ ) = P(Z > 1.52) = 1 - P(Z $\leq$ 1.52)

= 1 - 0.93574 = 0.0643

The above probability is calculated by looking at the value of x = 1.52 in the z table which gave an area of 0.93574.

(b) Probability that the worker’s annual salary is less than $46,000 is given by = P(X < $46,000)

P(X < $46,000) = P( $\frac{X-\mu}{\sigma }$ < $\frac{46,000-50,542}{4,246 }$ ) = P(Z < -1.07) = 1 - P(Z $\leq$ 1.07)

= 1 - 0.85769 = 0.1423

The above probability is calculated by looking at the value of x = 1.07 in the z table which gave an area of 0.85769.

(c) Probability that the worker’s annual salary is more than $40,000 is given by = P(X > $40,000)

P(X > $40,000) = P( $\frac{X-\mu}{\sigma }$ > $\frac{40,000-50,542}{4,246 }$ ) = P(Z > -2.48) = P(Z < 2.48)

= 1 - 0.99343 = 0.0066

The above probability is calculated by looking at the value of x = 2.48 in the z table which gave an area of 0.99343.

(d) Probability that the worker’s annual salary is between $45,000 and $54,000 is given by = P($45,000 < X < $54,000)

P($45,000 < X < $54,000) = P(X < $54,000) - P(X $\leq$ $45,000)

P(X < $54,000) = P( $\frac{X-\mu}{\sigma }$ < $\frac{54,000-50,542}{4,246 }$ ) = P(Z < 0.81) = 0.79103

P(X $\leq$ $45,000) = P( $\frac{X-\mu}{\sigma }$ $\leq$ $\frac{45,000-50,542}{4,246 }$ ) = P(Z $\leq$ -1.31) = 1 - P(Z < 1.31)

= 1 - 0.90490 = 0.0951

The above probability is calculated by looking at the value of x = 0.81 and x = 1.31 in the z table which gave an area of 0.79103 and 0.9049 respectively.

Therefore, P($45,000 < X < $54,000) = 0.79103 - 0.0951 = 0.6959

3 0

3 years ago