If there are 16 red bobbles for every 2 green dandees, how many red bobbles are there in 20 green dandees?

The number of bacteria, b, in a refrigerated food is given by the function b(t) = 15t2 − 60t + 125, where t is the temperature o

KonstantinChe [14]

Its an undefined value...

5 0

2 years ago

-6/5-2/3v+4/15+1/3v<br><br> What's the answer?

Georgia [21]

Answer:

You see the answer just character limit

7 0

2 years ago

Hii! I need some help, would anyone mind helping me?

Kisachek [45]

Step-by-step explanation:

${x}^{2} - 14x - 25 {y}^{2} + 100y = 76$

Factor by grouping,

$( {x}^{2} - 14x) - (25 {y}^{2} - 100y) = 76$

Complete the square, with the x variables,

$( {x}^{2} - 14x + 49) - (25 {y}^{2} - 100y) = 125$

Factor out 25 for the y variables

$( {x}^{2} - 14x + 49) - 25( {y}^{2} - 4y) = 125$

Complete the square

$( {x}^{2} - 14x + 49) - 25( {y}^{2} - 4y + 4) = 25$

Simplify the perfect square trinomial

$(x - 7) {}^{2} - 25(y - 2) {}^{2} = 25$

Make the right side be 1 so divide everything by 25.

$\frac{(x - 7) {}^{2} }{25} - (y - 2) {}^{2} = 1$

Here our center is (7,2).

8 0

2 years ago

A professor would like to test the hypothesis that the average number of minutes that a student needs to complete a statistics e

professor190 [17]

Answer:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

Step-by-step explanation:

Information given

$n=15$ represent the sample size

$\alpha=0.05$ represent the confidence level

$s^2 =16$ represent the sample variance

$\sigma^2_0 =25$ represent the value that we want to verify

System of hypothesis

We want to test if the true deviation for this case is lesss than 5minutes, so the system of hypothesis would be:

Null Hypothesis: $\sigma^2 \geq 25$

Alternative hypothesis: $\sigma^2$

The statistic is given by:

$\chi^2 =\frac{n-1}{\sigma^2_0} s^2$

And replacing we got:

$\chi^2 =\frac{15-1}{25} 16 =8.96$

The degrees of freedom are given by:

$df = n-1 = 15-1=14$

The p value for this case would be given by:

$p_v =P(\chi^2$

Since the p value is higher than the significance level we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviation is not ignificantly lower than 5 minutes

6 0

3 years ago

for a March past competition on sports day the house captain of one house cut 25 pieces of ribbon each 1/2 metre in length.what

zzz [600]

Answer:

not lenguague spanish

Step-by-step explanation:

that bad

7 0

3 years ago