Solve the inequality -4x>32=

In 1928, when the high jump was first introduced as a women's sport at the Olympic Games, the winning jump for women was 70.0 in

sveta [45]

Answer: year 2194

Explanation:

1) Model the jump of the women:

i) initial jump = 70.0 in

ii) increase rate = 0.48% per year

iii) equation: h = 70.0(1 +0.0048)ⁿ

2) Model the jump of the men:

i) initial jump 86.5 inches.

ii) increase rate, 0.4%.

iii) equation: h = 86.5 (1 + 0.004)ⁿ

3) Equal the two equations (h = h) to find when the jumps will be equal:

i) 70.0(1 +0.0048)ⁿ = 86.5 (1 + 0.004)ⁿ

ii) 70.0(1.0048)ⁿ = 86.5 (1.004)ⁿ

ii) [1.0048ⁿ / 1.004ⁿ] = 86.5/70.0

iii) [1.0048/ 1.0040]ⁿ = 86.5/70.0

iv) n log (1.0048 / 1.0040) = log ( 86.5/70)

v) n = log (86.5/70) / log (1.0048/1.0040) ≈ 266

3) year = 1928 + 266 = 2194

That is in the year 2194

8 0

3 years ago

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Please help me fill this in

Rom4ik [11]

Answer:

380.13m squared

you take half of the diameter for the radius and you do 11 squared times pie

7 0

3 years ago

A powder diet is tested on 49 people, and a liquid diet is tested on 36 different people. Of interest is whether the liquid diet

brilliants [131]

Answer:

There is not enough evidence to support the claim that the liquid diet yields a higher mean weight loss than the powder diet (P-value = 0.15).

Step-by-step explanation:

This is a hypothesis test for the difference between populations means.

The claim is that the liquid diet yields a higher mean weight loss than the powder diet.

Then, the null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a:\mu_1-\mu_2< 0$

The significance level is 0.05.

The sample 1 (powder diet group), of size n1=49 has a mean of 42 and a standard deviation of 12.

The sample 2 (liquid diet group), of size n2=36 has a mean of 45 and a standard deviation of 14.

The difference between sample means is Md=-3.

$M_d=M_1-M_2=42-45=-3$

The estimated standard error of the difference between means is computed using the formula:

$s_{M_d}=\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}=\sqrt{\dfrac{12^2}{49}+\dfrac{14^2}{36}}\\\\\\s_{M_d}=\sqrt{2.939+5.444}=\sqrt{8.383}=2.895$

Then, we can calculate the t-statistic as:

$t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-3-0}{2.895}=\dfrac{-3}{2.895}=-1.04$

The degrees of freedom for this test are:

$df=n_1+n_2-1=49+36-2=83$

This test is a left-tailed test, with 83 degrees of freedom and t=-1.04, so the P-value for this test is calculated as (using a t-table):

$\text{P-value}=P(t$

As the P-value (0.15) is greater than the significance level (0.05), the effect is not significant.

The null hypothesis failed to be rejected.

There is not enough evidence to support the claim that the liquid diet yields a higher mean weight loss than the powder diet.

6 0

3 years ago

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Which situation can be represented by Y= 12x - 4?

andrey2020 [161]

Answer:

D. The number of eggs , y, in x, dozen eggs for sale after 4 dozen eggs are sold

Step-by-step explanation:

The correct answer is D

The number of eggs , y, in x, dozen eggs for sale after 4 dozen eggs are sold.

Hope this helps.

4 0

2 years ago

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The graph below represents the solution set of which inequality?

natulia [17]

Answer:

option: B ( $x^2+2x-8$ ) is correct.

Step-by-step explanation:

We are given the solution set as seen from the graph as:

(-4,2)

1)

On solving the first inequality we have:

$x^2-2x-8$

On using the method of splitting the middle term we have:

$x^2-4x+2x-8$

⇒ $x(x-4)+2(x-4)=0$

⇒ $(x+2)(x-4)$

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

$x+2>0$ and $x-4$

i.e. x>-2 and x<4

so we have the region as:

(-2,4)

Case 2:

$x+2$ and $x-4>0$

i.e. x<-2 and x>4

Hence, we did not get a common region.

Hence from both the cases we did not get the required region.

Hence, option 1 is incorrect.

2)

We are given the second inequality as:

$x^2+2x-8$

On using the method of splitting the middle term we have:

$x^2+4x-2x-8$

⇒ $x(x+4)-2(x+4)$

⇒ $(x-2)(x+4)$

And we know that the product of two quantities are negative if either one of them is negative so we have two cases:

case 1:

$x-2>0$ and $x+4$

i.e. x>2 and x<-4

Hence, we do not get a common region.

Case 2:

$x-2$ and $x+4>0$

i.e. x<2 and x>-4

Hence the common region is (-4,2) which is same as the given option.

Hence, option B is correct.

3)

$x^2-2x-8>0$

On using the method of splitting the middle term we have:

$x^2-4x+2x-8>0$

⇒ $x(x-4)+2(x-4)>0$

⇒ $(x-4)(x+2)>0$

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

$x+2>0$ and $x-4>0$

i.e. x>-2 and x>4

Hence, the common region is (4,∞)

Case 2:

$x+2$ and $x-4$

i.e. x<-2 and x<4

Hence, the common region is: (-∞,-2)

Hence, from both the cases we did not get the desired answer.

Hence, option C is incorrect.

4)

$x^2+2x-8>0$

On using the method of splitting the middle term we have:

$x^2+4x-2x-8>0$

⇒ $x(x+4)-2(x+4)>0$

⇒ $(x-2)(X+4)>0$

And we know that the product of two quantities are positive if either both of them are negative or both of them are positive so we have two cases:

Case 1:

$x-2$ and $x+4$

i.e. x<2 and x<-4

Hence, the common region is: (-∞,-4)

Case 2:

$x-2>0$ and $x+4>0$

i.e. x>2 and x>-4.

Hence, the common region is: (2,∞)

Hence from both the case we do not have the desired region.

Hence, option D is incorrect.

5 0

3 years ago