Whats the value of -6

A silver decoration of width 4 cm runs immediately inside the boundary of a photo frame of size 28 cm multiply 24 cm.

konstantin123 [22]

Answer:

gebdododkdbdvscsbdha bsjsjsos

5 0

2 years ago

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Given the following exponential function, identify whether the change represents growth or decay, and determine the percentage r

saw5 [17]

Using exponential function concepts, it is found that it represents a decay of 0.9%.

<h3>What is an exponential function?</h3>

A decaying exponential function is <em>modeled </em>by:

$A(t) = A(0)(1 - r)^t$

In which:

A(0) is the initial value.

r is the decay rate, as a decimal.

In this problem, the function is:

$A(t) = 9300(0.991)^{6t}$

Hence:

$1 - r = 0.991$

$r = 0.009$

Thus, it represents a decay of 0.9%.

You can learn more about exponential function concepts at brainly.com/question/25537936

5 0

2 years ago

The sum of t and 2 is equal to 5 less than t

Alex17521 [72]

The sum (addition) of t and 2 is equal to (=) 5 less than (subtraction) t

2 + t = t - 5

7 0

3 years ago

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An investigator thinks that people under the age of forty have vocabularies that are different than those of people over sixty y

Kay [80]

Answer:

$t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28$

Now we can calculate the p value with the following probability:

$p_v =2*P(t_{60}$

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

Step-by-step explanation:

Information given

$\bar X_{1}=14$ represent the mean for sample 1 (younger)

$\bar X_{2}=20$ represent the mean for sample 2 (older)

$s_{1}=5$ represent the sample standard deviation for 1

$s_{f}=6$ represent the sample standard deviation for 2

$n_{1}=31$ sample size for the group 2

$n_{2}=31$ sample size for the group 2

t would represent the statistic

System of hypothesis

We want to test if that people under the age of forty have vocabularies that are different than those of people over sixty years of age, the system of hypothesis are:

Null hypothesis: $\mu_{1}-\mu_{2}=0$

Alternative hypothesis: $\mu_{1} - \mu_{2}\neq 0$

The statistic is given by:

$t=\frac{(\bar X_{1}-\bar X_{2})-\Delta}{\sqrt{\frac{\sigma^2_{1}}{n_{1}}+\frac{\sigma^2_{2}}{n_{2}}}}$ (1)

And the degrees of freedom are given by $df=n_1 +n_2 -2=31+31-2=60$

Replacing the info given we got:

$t=\frac{(14-20)-0}{\sqrt{\frac{5^2}{31}+\frac{6^2}{31}}}}=-4.28$

Now we can calculate the p value with the following probability:

$p_v =2*P(t_{60}$

The p value is a very low value so then we have enough evidence to reject the null hypothesis and we can conclude that people under the age of forty have vocabularies that are different than those of people over sixty years of age.

6 0

3 years ago

What is the value of tan(60°)? One-half StartRoot 3 EndRoot StartFraction StartRoot 3 EndRoot Over 2 EndFraction StartFraction 1

Ratling [72]

Answer:

$\sqrt3$