Solve equation 30t+8=20t+98

75 POINTS Which equations are equivalent to y = two-thirds x minus 4 when written in slope-intercept form? Check all that apply.

JulsSmile [24]

Answer:

xx

Step-by-step explanation:

5 1

2 years ago

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Model Exponential Growth Question :A sample of bacteria is growing at a continuously compounding rate. The sample triples in 10

Iteru [2.4K]

Answer:

The number of bacteria $B$ after $d$ days is given by

$B = B_0 (3)^{\frac{1}{10} d}$

where $B_0$ is the initial number of bacteria.

Step-by-step explanation:

The number of bacteria $B$ in the sample triples every 10 days, this means after the first 10th day, the number of bacteria is

$B = B_0 *3,$

where $B_0$ is the initial number of bacteria in the sample.

After the 2nd 10th days, the number of bacteria is

$B = (B_0 *3)*3$

after the 3rd day,

$B =( B_0 *3*3)*3$

and so on.

Thus, the formula we get for the number of bacteria after the <em>n</em>th 10-days is

$B = B_0 (3)^n$

where $n$ is is the <em>n</em>th 10-days.

Since, $n$ is 10 days, we have

$d =10n$

or

$n =\dfrac{1}{10}$

Substituting that into $B = B_0 (3)^n$ , we get:

$\boxed{ B = B_0 (3)^{\frac{1}{10} d}}$

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

a player gets to throw 4 darts at the target shown. Assuming the player will always hit the target, the probability of hitting a

padilas [110]

Answer:

3.375

Step-by-step explanation:

(3/5)^3 / (2/5)^3 = 3.375

3 0

2 years ago

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A shark began at 172.5 meters below sea level and then swam up 137.1 meters where is the sharks location now relation to sea lev

shutvik [7]

Answer:34.4

Step-by-step explanation:

172.5-137.1=34.4

8 0

2 years ago