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4 years ago

For exercise, mr.harris walked 1 and 3 fourths miles 5 times last week.how many miles did he walk in all?

Mathematics

2 answers:

Digiron [165]4 years ago

8 0

1(3/4) miles 5 times= 1(3/4)*5= 8 and 3/4 miles

RUDIKE [14]4 years ago

4 0

Well, if he walked one and three fourths miles five times, all you have to do is multiply one and three fourths and five to get your answer.

$1 \frac{3}{4}*5 \\ \\ \frac{7}{4}* \frac{5}{1} \\ \\ \frac{7*5}{4*1} \\ \\ \frac{35}{4} =8 \frac{3}{4}$

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The wildlife department has been feeding a special food to rainbow trout fingerlings in a pond. based on a large number of obser

Liono4ka [1.6K]

The normal distribution curve for the problem is shown below

We need to standardise the value X=405.5 by using the formula

$z-score= \frac{X-Mean}{Standard Deviation}$
$z-score= \frac{405.5-402.7}{8.8} =0.32$

We now need to find the probability of z=0.32 by reading the z-table

Note that z-table would give the reading to the left of z-score, so if your aim is to work out the area to the right of a z-score, then you'd need to do:

$P(Z\ \textgreater \ z)=1 - P(Z\ \textless \ z)$

from the z-table, the reading $P(Z\ \textless \ 0.32)$ gives 0.6255

hence,
$P(Z\ \textgreater \ 0.32)=1-0.6255=0.3475$

The probability that the mean weight for a sample of 40 trout exceeds 405.5 gram is 0.3475 = 34.75%

7 0

3 years ago

A poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ag

Katyanochek1 [597]

Answer:

At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll

Step-by-step explanation:

Given that a poll was taken this year asking college students if they considered themselves overweight. A similar poll was taken 5 years ago.

Let five years ago be group I X and as of now be group II Y

$H_0: p_x =p_y\\H_a: p_x \neq p_y$

(Two tailed test at 5% level of significance)

Group I Group II combined p

n 270 300 570

favor 120 140 260

p 0.4444 0.4667 0.4561

Std error for differene = $\sqrt{\frac{0.4561(1-0.4561)}{570} } \\=0.0209$

p difference = -0.0223

Z statistic = p diff/std error = -1.066

p value =0.2864

Since p value >0.05, we accept null hypothesis.

At 5% significance level, it is statistically evident that there is nodifference in the proportion of college students who consider themselves overweight between the two poll

7 0

3 years ago

In the seventh grade, Rachel read 15 books.

andre [41]

Answer:

20%

Step-by-step explanation:

Calculate percentage change

from V1 = 15 to V2 = 18

(V2−V1)|V1|×100

=(18−15)|15|×100

=315×100

=0.2×100

=20%change

=20%increase

4 0

3 years ago

Read 2 more answers

Simplify the following expression<br> 12b+5.54b+2.46b+3.2a+1.17a-2b

alekssr [168]

Answer:

Step-by-step explanation:

1. First you add like terms

12b+5.54b+2.46b-2b = 18b

3.2a+1.17a = 4.37a

2. Now you write it out

18b+4.37a

3. CELEBRATE!!!!

4. Pat yourself on the back one more problem done

5. Have a nice day

6 0

3 years ago

. Suppose you plan to make a box that will hold exactly 40 one-inch cubes. Give the dimensions (ex: 1in x 1in x 40in) of 3 possi

stiks02 [169]

Answer:

The dimensions of the three possible boxes are;

Box 1: 2 inches length by 4 inches width by 5 inches height

Box 2: 2 inches length by 2 inches width by 10 inches height

Box 3: 8 inches length by 1 inch width by 5 inches height

Step-by-step explanation:

The dimensions of the required box = 40 in³

Therefore, a box that will hold all the cubes will have a dimension given as follows;

Length, l × Width, w × Height, h = 40 in.³

We therefore, look for possible dimensions of l, w, h, however to make is faster, we note that l × w = base area, BA, hence we look for the factors 40 thus;

BA × h ≡ l × w × h

8 × 5 = 40 ≡ 2 × 4 × 5 = 40

4 × 10 = 40 ≡ 2 × 2 × 10 = 40

8 × 5 = 40 ≡ 8 × 1 × 5 = 40

Hence we have our three possible boxes as follows;

Box 1: Length = 2 inches, width = 4 inches and height = 5 inches

Box 2: Length = 2 inches, width = 2 inches and height = 10 inches

Box 3: Length = 8 inches, width = 1 inches and height = 5 inches.

4 0

3 years ago