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hammer [34]
2 years ago
8

The student population at Oak Mountain High School for a given year can be modeled by the function P(t)=550(1.12)^t , where t is

the number of years since 1995. How many students were there in 1995?
Mathematics
1 answer:
Anton [14]2 years ago
3 0

Answer:

There were 550 students in the school in the year 1995

Step-by-step explanation:

In this question, given the exponential equation, we are asked to calculate the number of students in a particular year.

The exponential equation expected to be used is given as;

P(t) = 550(1.12)^t

Now we are told that t is the number of years since 1995 and we are told to calculate the number of students in that same year 1995. Thus, the value of t would be 1995-1995 = 0

We substitute this value of t into the equation;

P(0) = 550(1.12)^0

Since anything raised to the power of zero is 1, then ;

P(0) = 550 * 1 = 550

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Using approximations to 1 significant figure,
malfutka [58]

Answer:

200

Step-by-step explanation:

Given:

0.482 x 61.2^2 ÷ √98.01

61.2^2 = 3745.44

√98.01 = 9.9

So,

0.482 × 3745.44 ÷ 9.9

= 1,805.30208 ÷ 9.9

= 182.35374545454

To one significant figure

= 200

One significant figure means only 1 non zero value and others are zero

5 0
2 years ago
If the 180 food calories found in a granola bar come entirely from p grams of protein, f grams of fat, and c grams of carbohydra
Mice21 [21]

f = 180 - p - c

to find the amount of fat subtract the protein and carbs from the original 180 calories

3 0
2 years ago
Read 2 more answers
Object A travels in the +x-direction before hitting a stationary object B. Afterwards, object AÍs x-momentum is 5.7 _ 104 kilogr
ANEK [815]
Below are the choices that can be found from other sources:

A. 42 
<span>B. 45 </span>
<span>C. 47 </span>
<span>D. 48 </span>
<span>E. 49
</span>
The answer is C or 47. The object’s resultant angle of motion with the +x-axis after the collision is  47. The reason for that is f<span>rom object A’s x-momentum is 5.7 × 104 kilogram meters/second and its y-momentum is 6.2 × 104 kilogram meters/second, we know that tan of the angle from the x-axis is 6.2 / 5.7 = 1.09 and acrtan 1.09 = 47.4</span>
8 0
2 years ago
Read 2 more answers
Tony is 4 years younger than his brother josh and two years older than his sister Cindy. Tony also has a twin brother, Evan. All
Sholpan [36]

Let Tony's age = x

He is 4 years younger than his brother Josh, so Josh's age would be x + 4

He is 2 years older than his sister, so her age would be x - 2

He has a twin, which would be the same age, so the twins age is also x

They all add together to equal 66, so you get:

x  + x + x+4 + x-2 = 66

Simplify:

4x +2 = 66

Subtract 2 from both sides:

4x = 64

Divide both sides by 4:

x = 64/4 = 16

Tony is 16 years old.

8 0
2 years ago
Two samples each of size 20 are taken from independent populations assumed to be normally distributed with equal variances. The
Harlamova29_29 [7]

Answer:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

\sigma^2_1 =\sigma^2_2 =\sigma^2

And the statistic is given by this formula:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

Where t follows a t distribution with n_1+n_2 -2 degrees of freedom and the pooled variance S^2_p is given by this formula:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

The system of hypothesis on this case are:

Null hypothesis: \mu_1 = \mu_2

Alternative hypothesis: \mu_1 \neq \mu_2

We have the following data given:

n_1 =20 represent the sample size for group 1

n_2 =20 represent the sample size for group 2

\bar X_1 =43.5 represent the sample mean for the group 1

\bar X_2 =40.1 represent the sample mean for the group 2

s_1=4.1 represent the sample standard deviation for group 1

s_2=3.2 represent the sample standard deviation for group 2

First we can begin finding the pooled variance:

\S^2_p =\frac{(20-1)(4.1)^2 +(20 -1)(3.2)^2}{20 +20 -2}=13.525

And the deviation would be just the square root of the variance:

S_p=3.678

The statistic is givne by:

t=\frac{(43.5 -40.1)-(0)}{3.678\sqrt{\frac{1}{20}+\frac{1}{20}}}=2.923

The degrees of freedom are

df=20+20-2=38

And the p value is given by:

p_v =2*P(t_{38}>2.923) =0.0058

Since the p value for this cae is lower than the significance level of 0.05 we have enough evidence to reject the null hypothesis and we can conclude that the true means for this case are significantly different

6 0
2 years ago
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