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2 years ago

It takes Fred 32 minutes to type and spell check 5 pages of a manuscript. Find how long it takes

Mathematics

1 answer:

sveta [45]2 years ago

6 0

Answer:

Step-by-step explanation:

$we \: rounded \: it \: to \: the \: nearest \: whole \: number$

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Dimension of a cuboid is 4 m 3 m ad 5 m how many such cuboid can form a cube....... please and urgent..experts please help

larisa [96]

Step-by-step explanation:

We dimension of the smallest cube to be made from cuboids of sides 3 m , 4 m and 5 m will be the least common multiple of 3 m, 4 m and 5 m i.e. 60 m

12 cuboidal should be stacked along 5 m edge to 60 m, 15 cuboids should be stacked along 4 m edge and 20. cuboids should be stacked along edge to make a cube of 60 m edge, Hence number of cuboids are 12× 15 ×20=3600

hope it helps

4 0

3 years ago

8. Identify the x and y intercept of the equation:<br> 12x - 8y = 48

GalinKa [24]

Answer:

X- intercept(S) : (4,0)

Y- Intercept(S) : (0, -6)

Step-by-step explanation:

To find the x-intercept, substitute in 0 for y and solve for x. To find the y-intercept, substitute in 0 for x and solve for Y.

Hopes this helps!

8 0

2 years ago

Need help ASAP. Please

Tanya [424]

I hope this helps you

53=y-127/2

answer is C

8 0

3 years ago

Read 2 more answers

One inch is equivalent to 2.54 centimeters. how many inches are in 50.8 cenemeters

777dan777 [17]

50.8 cm * 1 in/2.54 cm

50.8 / 2.54 inch

20 inches

3 0

3 years ago

Read 2 more answers

Toxaphene is an insecticide that has been identified as a pollutant in the Great Lakes ecosystem. To investigate the effect of t

Likurg_2 [28]

Answer:

This data suggest that there is more variability in low-dose weight gains than in control weight gains.

Step-by-step explanation:

Let $\sigma_{1}^{2}$ be the variance for the population of weight gains for rats given a low dose, and $\sigma_{2}^{2}$ the variance for the population of weight gains for control rats whose diet did not include the insecticide.

We want to test $H_{0}: \sigma_{1}^{2} = \sigma_{2}^{2}$ vs $H_{1}: \sigma_{1}^{2} > \sigma_{2}^{2}$ . We have that the sample standard deviation for $n_{2} = 22$ female control rats was $s_{2} = 28$ g and for $n_{1} = 18$ female low-dose rats was $s_{1} = 51$ g. So, we have observed the value

$F = \frac{s_{1}^{2}}{s_{2}^{2}} = \frac{(51)^{2}}{(28)^{2}} = 3.3176$ which comes from a F distribution with $n_{1} - 1 = 18 - 1 = 17$ degrees of freedom (numerator) and $n_{2} - 1 = 22 - 1 = 21$ degrees of freedom (denominator).

As we want carry out a test of hypothesis at the significance level of 0.05, we should find the 95th quantile of the F distribution with 17 and 21 degrees of freedom, this value is 2.1389. The rejection region is given by {F > 2.1389}, because the observed value is 3.3176 > 2.1389, we reject the null hypothesis. So, this data suggest that there is more variability in low-dose weight gains than in control weight gains.

8 0

3 years ago