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3 years ago

MathsWatch question please help me

Mathematics

1 answer:

nika2105 [10]3 years ago

5 0

Answer:

2.55°C

Step-by-step explanation:

Calculate the temperature at the 2 given depths

x = 1200

T = $\frac{4500}{1200}$ = 3.75

x = 3750

T = $\frac{4500}{3750}$ = 1.2

Difference in temperature = 3.75 - 1.2 = 2.55°C

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Analyze the reflectional symmetry of the letters.

Alona [7]

Answer:

C , T , I , O , S , H , K

Step-by-step explanation:

A 1-fold reflectional symmety means that if the letter is folded in the middle (either horizontally or vertically) then both sides will be identical and match up perfectly. From the words provided the letters that have this symmetry are the following.

From "RAINY" = I

From "ACTION" = C , T , I , O

From "SHARK" = S , H , K

Each of these letters can be folded once and have both parts identical to one another.

5 0

3 years ago

Given 1+ cos x/ sin x + sin x/1+ cos x= 4, find a numerical value of one trigonometric function of x.

laiz [17]

Answer:

The numerical value of the trigonometric function is 30 °

Step-by-step explanation:

Given trigonometric function as :

$\frac{1 + cos x}{sin x}$ + $\frac{sin x}{1 + cos x}$ = 4

or, Taking LCM we get

$\frac{(1+cosx)^{2}+sin^{2}x}{(sinx)\times (1+cosx)}$ = 4

Or, ( 1 + cos x )² + sin² x = 4 × ( sin x ) × ( 1 + cos x )

1 + cos² x + 2 cox + sin² x = 4 sin x + 4 sin x × cos x

or, ( cos² x + sin² x ) + ( 1 + 2 cos x ) = 4 sin x ( 1 + cos x )

∵ cos² x + sin² x = 1

or, 1 + 1 + 2 cos x = 4 sin x ( 1 + cos x )

or, 2 + 2 cos x = 4 sin x ( 1 + cos x )

or, 2 ( 1 + cos x ) = 4 sin x ( 1 + cos x )

Or, $\frac{2 ( 1+ cos x )}{4 ( 1 + cos x )}$ = sin x

Or, sin x = $\frac{1}{2}$

∴ x = $sin^{-1}\frac{1}{2}$

∵ sin 30 ° = $\frac{1}{2}$

I.e x = 30 °

Hence The numerical value of the trigonometric function is 30 ° answer

3 0

3 years ago

Jose, baulus, and George are having a disagreement over interest rates. Jose says that 6 3/4% can be expressed as 6.75%. George

otez555 [7]

Answer:

Jose and George are correct

Baulus is incorrect

Step-by-step explanation:

Given

$6\frac{3}{4}$ %

Required

Compare the above expression with

Jose: 6.75%.

George: 0.0675

Baulus: 0.0675%

The expression  $6\frac{3}{4}$ % is given as a mixed fraction;

In Jose's case, he simply converted it to decimal

Converting $6\frac{3}{4}$ % to decimal;

First, we need to convert to improper fraction

$6\frac{3}{4}$ % in improper fraction is $\frac{27}{4}$ %

Then, we divide the numerator by the denominator

$\frac{27}{4}$ % = 6.75%.

Hence, Jose is correct

In George's case, he solved further by removing the % from Jose's answer.

% means divided by 100; In other words, x% means that we divide x by 100

So, 6.75% means we divide 6.75 by 100

$\frac{6.75}{100}$ = 0.0675

Hence, George is also correct

But same can't be said of Baulus;

Baulus converted $6\frac{3}{4}$ % as 0.0675%

This means that he converted $6\frac{3}{4}$ as 0.0675 then he appended it with the percentage sign (%)

Since $6\frac{3}{4}$ is not equal to 0.0675, then we can conclude that Baulus is incorrect

8 0

4 years ago

Your class has 30 students. if 1/3 of them walk to school, how many students in your class walk to school

Delvig [45]

1/3 of 30 would be 10. So 10 students walk to school.

4 0

4 years ago

Three populations have proportions 0.1, 0.3, and 0.5. We select random samples of the size n from these populations. Only two of

IRINA_888 [86]

Answer:

(1) A Normal approximation to binomial can be applied for population 1, if n = 100.

(2) A Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3) A Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

Step-by-step explanation:

Consider a random variable X following a Binomial distribution with parameters n and p.

If the sample selected is too large and the probability of success is close to 0.50 a Normal approximation to binomial can be applied to approximate the distribution of X if the following conditions are satisfied:

np ≥ 10
n(1 - p) ≥ 10

The three populations has the following proportions:

p₁ = 0.10

p₂ = 0.30

p₃ = 0.50

(1)

Check the Normal approximation conditions for population 1, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.10=1$

Thus, a Normal approximation to binomial can be applied for population 1, if n = 100.

(2)

Check the Normal approximation conditions for population 2, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.30=310\\\\n_{c}p_{1}=50\times 0.30=15>10\\\\n_{d}p_{1}=40\times 0.10=12>10\\\\n_{e}p_{1}=20\times 0.10=6$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50 and 40.

(3)

Check the Normal approximation conditions for population 3, for all the provided n as follows:

$n_{a}p_{1}=10\times 0.50=510\\\\n_{c}p_{1}=50\times 0.50=25>10\\\\n_{d}p_{1}=40\times 0.50=20>10\\\\n_{e}p_{1}=20\times 0.10=10=10$

Thus, a Normal approximation to binomial can be applied for population 2, if n = 100, 50, 40 and 20.

8 0

3 years ago