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3 years ago

7.5 - 3.2 + 4³ ÷ 2I NEED HELP PLIS :(

Mathematics

2 answers:

scoray [572]3 years ago

7 0

Answer:

36.3

Step-by-step explanation:

Step 1: 7.5-3.2+2⁶÷2

Step 2: 7.5-3.2+2⁵

Step 3: 7.5-3.2+32

Step 4: 7.5-3.2+32

Gala2k [10]3 years ago

6 0

Answer:

36.3

Step-by-step explanation:

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Plsssssssssss Help!!!!!

Nuetrik [128]

Answer:

1. y=x^2+1 is nonlinear

2. y=x^2+2 is linear

3. y=x is linear

4. y=5(x+2) is linear

5. y=x(x) is nonlinear

hope this helps !!

Step-by-step explanation:

5 0

3 years ago

The game commission observes the fish population in a stream and notices that the number of trout increases by a factor of 1.5 e

just olya [345]

Answer:

A - $f(t)=80(1.5)^t$

B - Graph given below

C - Number of trouts in 5th week are 607.2

D - Population of trouts will exceed 500 on the 5th week.

Step-by-step explanation:

We are given that,

The number of trout increases by a factor of 1.5 each week and the initial population of the trout is observed to be 80.

Part A: So, the explicit formula representing the situation is,

$f(t)=80(1.5)^t$ , where f(t) represents the population of trouts after 't' weeks.

Part B: The graph of the function can be seen below.

It can be seen that the function is an exponential function.

Part C: It is required to find the number of trouts in the 5th week.

So, we have,

$f(5)=80(1.5)^5$

i.e. $f(5)=80\times 7.59$

i.e. f(5) = 607.2

Thus, the number of trouts in 5th week are 607.2

Part D: We are given that the trout population exceeds 500.

It is required to find the week in which this happens.

So, we have,

$50$

i.e. $(1.5)^t>\frac{500}{80}$

i.e. $(1.5)^t>6.25$

i.e. $t\log 1.5>\log 6.25$

i.e. $t\times 0.1761>0.7959$

i.e. $t>\frac{0.7959}{0.1761}$

i.e. t > 4.5

As, t represents the number of weeks. So, to nearest whole, t = 5.

Thus, the population of trouts will exceed 500 on the 5th week.

7 0

3 years ago

Read the questions and choose the best answer.

Arte-miy333 [17]

Where’s the questions ?

8 0

2 years ago

A basket contains four apples, three peaches, and four pears. You randomly select and eat three pieces of fruit. The first piece

Artist 52 [7]

We have 4 apples, 3 peaches, and 4 pears.
The sample size is 11.

1st selection: apple
The probability of selecting an apple is 4/11.

2nd selection: peach
We now have a sample size of 10, which consists of 3 apples, 3 peaches, and 4 pears.
The probability of selecting a peach is 3/10.

3rd selection: peach
We now have a sample size of 9, which consists of 3 apples, 2 peaches, and 4 pears.
The probability of selecting a peach is 2/9.

Each selection was an independent event. Therefore the overall probability is
(4/11)*(3/10)*(2/9) = 4/165

Answer:
The probability of selecting an apple followed by 2 peaches is 4/165.
This probability is equivalent to 0.024 or 2.4%.

8 0

4 years ago

Read 2 more answers

) How many degrees of freedom are available for hypothesis tests and confidence intervals for single regression coefficient

sladkih [1.3K]

Answer:

We use students' t distribution therefore degrees of freedom is v= n-2

Step-by-step explanation:

<u>Confidence Interval Estimate of Population Regression Co efficient β.</u>

To construct the confidence interval for β, the population regression co efficient , we use b, the sample estimate of β. The sampling distribution of b is normally distributed with mean β and a standard deviation σ.y.x / √(x-x`)². That is the variable z = b - β/σ.y.x / √(x-x`)² is a standard normal variable. But σ.y.x is not known so we use S.y.x and also student's t distribution rather than normal distribution.

t= b - β/S.y.x / √(x-x`)² = b - β/Sb [Sb = S.y.x / √(x-x`)²]

with v= n-2 degrees of freedom.

Consequently

P [ - t α/2< b - β/Sb < t α/2] = 1- α

P [ b- t α/2 Sb< β < b+ t α/2 Sb] = 1- α

Hence a 100( 1-α) percent confidence for β the population regression coefficient for a particular sample size n <30 is given by

b± t α/2 Sb

Using the same statistic a confidence interval for α can be constructed in the same way for β replacing a with b and Sa with Sb.

a± t α/2 Sa

Using the t statistic we may construct the confidence interval for U.y.x for the given value X0 in the same manner

Y~0 ± t α/2(n-2) SY~

Y~0= a+b X0

4 0

3 years ago

7.5 - 3.2 + 4³ ÷ 2I NEED HELP PLIS :(​

7.5 - 3.2 + 4³ ÷ 2I NEED HELP PLIS :(