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2 years ago

Whats double of 8x4?

Mathematics

2 answers:

Volgvan2 years ago

6 0

Answer:

Step-by-step explanation:

8x4 equals 32 but you said "double" so that would equal 64 unless you means 32^2 cause that would equal 1024. Have a great day!

Phoenix [80]2 years ago

5 0

Answer:

Step-by-step explanation:

cus 8 times 4 is 32 and 32 times 2 is 64

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Use the function below to find F-4). F(x) = 2x O A.1/8 O B.-16 O C.1/16 O D.-8

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Answer:

Option D is correct, -4 x 2x = 8x

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It would take the car 20 seconds

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Answer:

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Step-by-step explanation:

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2 years ago

The measurenicnt of the circumference of a circle is found to be 56 centimeters. The possible error in measuring the circumferen

BartSMP [9]

Answer:

(a) Approximate the percent error in computing the area of the circle: 4.5%

(b) Estimate the maximum allowable percent error in measuring the circumference if the error in computing the area cannot exceed 3%: 0.6 cm

Step-by-step explanation:

(a)

First we need to calculate the radius from the circumference:

$c=2\pi r\\r=\frac{c}{2\pi } \\c=8.9 cm$

I leave only one decimal as we need to keep significative figures

Now we proceed to calculate the error for the radius:

$\Delta r=\frac{dt}{dc} \Delta c\\\\\frac{dt}{dc} = \frac{1}{2 \pi } \\\\\Delta r=\frac{1}{2 \pi } (1.2)\\\\\Delta r= 0.2 cm$

$r = 8.9 \pm 0.2 cm$

Again only one decimal because the significative figures

Now that we have the radius, we can calculate the area and the error:

$A=\pi r^{2}\\A=249 cm^{2}$

Then we calculate the error:

$\Delta A= (\frac{dA}{dr} ) \Delta r\\\\\Delta A= 2\pi r \Delta r\\\\\Delta A= 11.2 cm^{2}$

$A=249 \pm 11.2 cm^{2}$

Now we proceed to calculate the percent error:

$\%e =\frac{\Delta A}{A} *100\\\\\%e =\frac{11.2}{249} *100\\\\\%e =4.5\%$

(b)

With the previous values and equations, now we set our error in 3%, so we just go back changing the values:

$\%e =\frac{\Delta A}{A} *100\\\\3\%=\frac{\Delta A}{249} *100\\\\\Delta A =7.5 cm^{2}$

Now we calculate the error for the radius:

$\Delta r= \frac{\Delta A}{2 \pi r}\\\\\Delta r= \frac{7.5}{2 \pi 8.9}\\\\\Delta r= 0.1 cm$

Now we proceed with the error for the circumference:

$\Delta c= \frac{\Delta r}{\frac{1}{2\pi }} = 2\pi \Delta r\\\\\Delta c= 2\pi 0.1\\\\\Delta c= 0.6 cm$

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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

2 years ago