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3 years ago

Write 14 in scientific notation

Mathematics

1 answer:

Lina20 [59]3 years ago

8 0

1.4 x 10^1 is the answer :)

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What is the lateral area of the drawing?

katrin2010 [14]

The lateral area of the shape is the sum of the area of all lateral faces.

The area of the lateral face that is rectangle with sides 5 mi and 2 mi is

$A=5\cdot 2=10\ mi^2.$

As you can see the shape has all four lateral faces of the same rectangular form with the same sides lengths, therefore,

$LA=4A=4\cdot 10=40\ mi^2.$

Answer: correct choice is B.

8 0

3 years ago

If it takes 6 hours for a plane to travel 720km with a tail wind and 8 hours to make the return trip with a head wind. Find the

denis-greek [22]

The speed of wind and plane are 105 kmph and 15 kmph respectively.

Solution:

Given, it takes 6 hours for a plane to travel 720 km with a tail wind and 8 hours to make the return trip with a head wind.

We have to find the air speed of the plane and speed of the wind.

Now, let the speed wind be "a" and speed of aeroplane be "b"

And, we know that, distance = speed x time.

$\text { Now, at tail wind } \rightarrow 720=(a+b) \times 6 \rightarrow a+b=\frac{720}{6} \rightarrow a+b=120 \rightarrow(1)$

Now at head wind → $720=(a-b) \times 8 \rightarrow a-b=\frac{720}{8} \rightarrow a-b=90 \rightarrow(2)$

So, solve (1) and (2) by addition

2a = 210

a = 105

substitute a value in (1) ⇒ 105 + b = 120

⇒ b = 120 – 105 ⇒ b = 15.

Here, relative speed of plane during tail wind is 120 kmph and during head wind is 90 kmph.

Hence, speed of wind and plane are 105 kmph and 15 kmph respectively.

6 0

3 years ago

Could someone help me understand how to solve this?

monitta

Answer:

Amount in 4% account: $550

Step-by-step explanation:

Use simple interest formula $I=P\cdot r\cdot t,$ where

I = interest

P = principal

r = rate (as decimal)

t = time.

4% account:

$P_1=x\\ \\r_1=0.04\\ \\t_1=1,$

then

$I_1=x\cdot 0.04\cdot 1=0.04x$

5% account:

$P_2=1,500-x\\ \\r_2=0.05\\ \\t_2=1,$

then

$I_2=(1,500-x)\cdot 0.05\cdot 1=0.05(1,500-x)$

You have earned $69.50 in total, so

$I_1+I_2=69.50\\ \\0.04x+0.05(1,500-x)=69.50\\ \\4x+5(1,500-x)=6,950\\ \\4x+7,500-5x=6,950\\ \\4x-5x=6,950-7,500\\ \\-x=-550\\ \\x=550\\ \\1,500-x=1,500-550=950$

Amount in 4% account: $550

Amount in 5% account: $950

8 0

3 years ago

Suppose the test scores for a college entrance exam are normally distributed with a mean of 450 and a s. d. of 100. a. What is t

svet-max [94.6K]

Answer:

a) 68.26% probability that a student scores between 350 and 550

b) A score of 638(or higher).

c) The 60th percentile of test scores is 475.3.

d) The middle 30% of the test scores is between 411.5 and 488.5.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 450, \sigma = 100$

a. What is the probability that a student scores between 350 and 550?

This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So

X = 550

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{550 - 450}{100}$

$Z = 1$

$Z = 1$ has a pvalue of 0.8413

X = 350

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{350 - 450}{100}$

$Z = -1$

$Z = -1$ has a pvalue of 0.1587

0.8413 - 0.1587 = 0.6826

68.26% probability that a student scores between 350 and 550

b. If the upper 3% scholarship, what score must a student receive to get a scholarship?

100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88

$Z = \frac{X - \mu}{\sigma}$

$1.88 = \frac{X - 450}{100}$

$X - 450 = 1.88*100$

$X = 638$

A score of 638(or higher).

c. Find the 60th percentile of the test scores.

X when Z has a pvalue of 0.60. So it is X when Z = 0.253

$Z = \frac{X - \mu}{\sigma}$

$0.253 = \frac{X - 450}{100}$

$X - 450 = 0.253*100$

$X = 475.3$

The 60th percentile of test scores is 475.3.

d. Find the middle 30% of the test scores.

50 - (30/2) = 35th percentile

50 + (30/2) = 65th percentile.

35th percentile:

X when Z has a pvalue of 0.35. So X when Z = -0.385.

$Z = \frac{X - \mu}{\sigma}$

$-0.385 = \frac{X - 450}{100}$

$X - 450 = -0.385*100$

$X = 411.5$

65th percentile:

X when Z has a pvalue of 0.35. So X when Z = 0.385.

$Z = \frac{X - \mu}{\sigma}$

$0.385 = \frac{X - 450}{100}$

$X - 450 = 0.385*100$

$X = 488.5$

The middle 30% of the test scores is between 411.5 and 488.5.

7 0

3 years ago

A bowl contained 250 skittles. The result of 25 fruit-flavored candies pulled at random is given below. Based on this informatio

Anastaziya [24]

OK so you have 250 skittles. But you only pull out 25. You pull out 5 red skittles, 7 yellow skittles,8 orange skittles,2 green skittles and 3 purple skittles. So how many red skittles are in the whole box. The way i would look at it is theres 5 red skittles in every 25 skittles. So whats 250 divided by 25 that would be 10. So you can imagine you have 10 groups of skittles. In each group theres 5 skittles. So all together there would be 50 red skittles. Sooo based on the infromation there would be 50 red skittles expected in the box. I hope this helped pls mark me as brainliest!!

7 0

2 years ago