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puteri [66]
4 years ago
11

Sarah hopes to be 5 feet tall by her next birthday. Right now, she is 148 centimeters tall. About how many more centimeters does

she need to grow before her next birthday to reach 5 feet? Use the drop-down menus to explain how to solve the problem. the number of inches in 5 ft by 2.54 cm. There are about cm in 5 ft. Then to find the number of centimeters that Sarah still needs to grow. She needs to grow cm to reach 5 ft.
Mathematics
1 answer:
Alla [95]4 years ago
3 0

Answer:

4.4 cm

Step-by-step explanation:

To solve this problem, we have to remind the conversion factor between feet and cm. We know that:

1 foot = 12 inches

1 inches = 2.54 cm

So,

1 ft = 12 \cdot 2.54 =30.48 cm

In this problem, Sarah's height right now is

h = 148 cm

She wants to be 5 feet tall by her next birthday: it means that her height should be

h'=5 ft \cdot 30.48 =152.4 cm

Therefore, she needs to grow by:

h'-h=152.4-148=4.4 cm

So, she needs to grow 4.4 cm more in order to be 5 feet tall.

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A line with points (-4.0) and (-3.1)<br> has a slope of?
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Slope is the change in y over the change in x

Slope = (1-0) /( -3 - -4)

Slope = 1/1

Slope = 1

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3 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
3 years ago
What is the sum?
Nezavi [6.7K]

Answer:

Its B on e2020

Step-by-step explanation:

correct on quiz

4 0
3 years ago
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A circle has a radius of 2.5 cm. What is the diameter of this circle in cm? Record your answer in Blank 1. What is the circumfer
Eva8 [605]

Answer:

The diameter is 5 cm and the circumference is diameter x pi

Step-by-step explanation:

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3 years ago
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poizon [28]

Answer:

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Step-by-step explanation:

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x is the number of feet away from the sprinkler head the spray

To get the height of the spray 2 feet away from the sprinkler head, we will simply substitute x =2 into the function and et the height h as shown;

From the equation

ℎ(x)=160−16x^2

h(2) = 160-16(2)²

h(2) = 160-16(4)

h(2) = 160-64

h(2) = 96feet

Hence the height will be 96feet if the spray is 2feet away from the sprinklers head

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