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torisob [31]
2 years ago
13

The weight of a soccer ball is normally distributed with a mean of 21 oz and a standard deviation of 3 oz. Suppose 1000 differen

t soccer balls are in a warehouse. About how many soccer balls weigh more than 24 oz
A. 40

B. 80

C. 160

D. 200
Mathematics
2 answers:
horsena [70]2 years ago
8 0

Answer:

The correct option is C.

Step-by-step explanation:

Given information: The weight of a soccer ball is normally distributed with a mean of 21 oz and a standard deviation of 3 oz. Total number of soccer balls is 1000.

We have to find the number of soccer balls having weight more than 24 oz.

Probability of the ball having weight more than 24 oz is

P(x>24)=P(\frac{x-\mu}{\sigma}>\frac{24-21}{3})

P(x>24)=P(z>1)

P(x>24)=1-P(z\leq 1)

P(x>24)=1-P(z\leq 1)                  (Using standard normal table)

P(x>24)=1-0.8413

P(x>24)=0.1587

The number of soccer balls having weight more than 24 oz is

1000\times P(x>24)=1000\times 0.1587\Rightarrow 158.7\approx 160

The number of soccer balls having weight more than 24 oz is about 160.

Therefore the correct option is C.

Novosadov [1.4K]2 years ago
3 0

A weight of 24 oz is 1 standard deviation above the mean 21 oz. The empirical rule says approximately 68% of soccer balls have weights within 1 standard deviation (i.e. between 18 and 24 oz), so the remaining 32% of balls have weights below 18 oz or above 24 oz. The balls' weights are normally distributed, so the percentage of balls with weight less than 18 oz is the same as the percentages of balls with weight greater than 24 oz, which means about 16% of balls have weight greater than 24 oz. 16% of 1000 is 160.

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A. (Assume both planets are aligned with the sun and are on the same side of the sun.) Show your work and provide your answer in scientific notation. Jupiter = 778,500,000 km from sun. Venus = 108,200,000 km from sun.

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10 {}^{8}


B.Mercury, Venus, and Earth are the three planets closest to the sun. Would their combined distance from the sun be greater or less than the distance from the sun to Neptune? Show your work and justify your answer. Mercury = 57,909,000 km Venus = 108,200,000 km Earth = 149,600,000 km 57,909,000 + 108,200,000 + 149,600,000 = 315,709,000 Less because 315,709,000 km is less than 4,498,000,000 km

C.If Earth was 10 times farther away from the sun than it is now, which planet would it be closest to? (Assume all the planets are aligned with the sun and are on the same side of the sun.) Compare Earth's new distance to that planet. How far apart would they be in standard notation? How far apart in scientific notation? Show your work. Earth = 149,600,000 km 149,600,000 x 10 = 1,496,000,000 Earth’s new distance = 1,496,000,000 Earth would be closest to Saturn. 1,496,000,000 – 1,429,000,000 = 67,000,000
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4 0
2 years ago
Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute
valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

       = 3 - 1 = 2

d.f.D = N - k

        = 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust                      Jet red                 Cloudtran

$\overline X_1 =50.5$           $\overline X_2 =50.07143$        $\overline X_3 =55.71429$

$s_1^2=19.96154$      $s_2^2=14.68681$         $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

         $=\frac{51+51+...+49+49}{35}$

        = 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

     $=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

   $=\frac{127.1143}{2}$

   = 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

    $=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

   $=\frac{669.8571}{32}$

  = 20.93304

The F-test  statistics value is :

$F=\frac{s_B^2}{s_W^2}$

  $=\frac{63.55714}{20.93304}$

  = 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source       Sum of squares    d.f    Mean square    F

Between    127.1143                 2      63.55714          3.036212

Within        669.8571             32      20.93304

Total           796.9714            34

3 0
1 year ago
What are the solutions to the equation 0 = x2 – x –6? Select two options
KatRina [158]

Answer:

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Step-by-step explanation:

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Since this is a quadratic all we have to do is find two numbers that multiply to be c and add up to be b to factor the expression that is to the left of the equal sign.

By the way a quadratic expression looks like ax^2+bx+c.

So we want to find two numbers that multiply to be -6 and add up to be -1.

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So this means we have the following two equations to solve:

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First equation we will add 3 on both sides.

Second equation we will subtract 2 on both sides.

x=3 or x=-2

6 0
2 years ago
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