Which expression represents the factors of 2x^2 + 7x + 6

Change the percent to a decimal.<br> 67% =

alekssr [168]

67% as a decimal would be .67. This is because, when you divide 67 over 100, you get .67. hope this helped!

7 0

3 years ago

Kaitlyn is walking on a treadmill at a constant pace for 28 minutes. She has programmed the treadmill for a 2-mile walk. The dis

maw [93]

Is this multiplying or what??? Anyways try calculating those numbers so basically you need to calculate 28x2 or if you have to divide them or multiply them you would get the answer is very simple

6 0

2 years ago

If the side lengths of a cube are 14 feet, what is the correct way to write the expression to represent the volume of the cube i

Kryger [21]

Answer: Choice A) $14^3$

This is the same as writing 14^3

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Explanation:

Since all the sides are 14 feet long, we can basically say length = width = height = 14.

Or put another way:

length = 14
width = 14
height = 14

The volume of the cube is length*width*height = 14*14*14 = 14^3 = 2744 cubic feet.

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The term 14^3 has a base of 14 and exponent of 3.

The base is what we multiply out while the exponent tells us how many copies of the base to multiply. In this problem, we have 3 copies of 14 being multiplied. So that's why 14^3 = 14*14*14

If you were to say something like 3^14, then you would have 14 copies of 3 multiplied out, which is a much larger value. So we'll cross choice B off the list.

Choice C isn't correct either since again we're using exponents and not multiplication to tie together the 14 and 3.

Choice D would be correct if your teacher wanted it in expanded form, instead of exponent form.

6 0

3 years ago

. Predict the weather that will be at your house when the front has arrived.

FrozenT [24]

As the front approaches, you will have a storm.

I hope this helps!

3 0

3 years ago

A dataset lists full IQ scores for a random sample of subjects with low lead levels in their blood (sample 1) and another random

Alenkasestr [34]

Answer:

a. Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

b. $t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

c. $p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

Step-by-step explanation:

a. State and label the null and alternative hypotheses.

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 \leq \mu_2$

Alternative hypothesis: $\mu_1 >\mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 \leq 0$

Alternative hypothesis: $\mu_1 -\mu_2>0$

Our notation on this case :

$n_1 =78$ represent the sample size for group 1

$n_2 =21$ represent the sample size for group 2

$\bar X_1 =92.88$ represent the sample mean for the group 1

$\bar X_2 =86.90$ represent the sample mean for the group 2

$s_1=15.34$ represent the sample standard deviation for group 1

$s_2=8.99$ represent the sample standard deviation for group 2

b. State the value of the test statistic.

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{\sqrt{\frac{s^2_1}{n_1}}+\frac{s^2_2}{n_2}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom. If we replace the values given we have:

$t=\frac{(92.88 -86.90)-(0)}{\sqrt{\frac{15.34^2}{78}}+\frac{8.99^2}{21}}=2.282$

Now we can calculate the degrees of freedom given by:

$df=78+21-2=97$

c. Find either the critical value(s) and draw a picture of the critical region(s) or find the P-value for this test. Indicate which method you are using: ( CIRCLE ONE: Critical value / P-value )

Method used: P value

And now we can calculate the p value using the altenative hypothesis, since it's a right tail test the p value is given by:

$p_v =P(t_{97}>2.287) =0.0122$

So with the p value obtained and using the significance level given $\alpha=0.05$ we have $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 5% of significance the mean of the group 1 (Low Blood Lead level) is significantly higher than the mean for the group 2 (High Blood Lead level).

6 0

3 years ago