All categories

2 years ago

Seven times the product of negative six and a number

Mathematics

2 answers:

Lynna [10]2 years ago

5 0

Hi!

I can help you with joy! :)

7 times the product of -6 and a number. Let the number be

a. Now, "the product of" means we multiply.

Multiply -6 times a: -6a

If you notice, we write the number before the variable.

Now, multiply 7 times -6a:

7(-6a)

Simplify:

-42a

I hope this helps!

Have a Great Day!

-Content Girl

$\bf{Mysterious^a^n^d\:M^a^gi^ca^l$

jonny [76]2 years ago

4 0

7 times -6x, because “a number” can be replaced with x

You might be interested in

What is the area where if this figurine ?ixl question

nydimaria [60]

Answer:

The area of the figure is 97 square inches

Step-by-step explanation:

In the given figure

To find the area of the figure we can find the area of the big rectangle with dimensions 7 inches and 19 inches and subtract from it the area of the small rectangle with dimensions 4 inches and 9 inches

∵ The area o the rectangle = length × width

∵ The dimensions of the big rectangle are 7 inches and 19 inches

∴ The width = 7 inches

∴ The length = 19 inches

→ Substitute them in the rule of the area above

∵ The area of the big rectangle = 19 × 7

∴ The area of the big rectangle = 133 inches²

∵ The dimensions of the small rectangle are 4 inches and 9 inches

∴ The width = 4 inches

∴ The length = 9 inches

→ Substitute them in the rule of the area above

∵ The area of the small rectangle = 9 × 4

∴ The area of the small rectangle = 36 inches²

→ Subtract the area of the small rectangle from the area of the big

rectangle to get the area of the green figure

∵ The area of the figure = the big area - the small area

∴ The area of the figure = 133 - 36

∴ The area of the figure = 97 inches²

∴ The area of the figure is 97 square inches

6 0

3 years ago

The mean time it takes to walk to the bus stop is 8 minutes (with a standard deviation of 2 minutes) and the mean time it takes

kifflom [539]

We have been given for a normal distribution the mean time it takes to walk to the bus stop is 8 minutes with a standard deviation of 2 minutes. And the mean time it takes for the bus to get to school is 20 minutes with a standard deviation of 4 minutes.

(a) Average time that it would take reach school can be obtained by adding the average times.

8+20 = 28 minutes.

(b) Standard deviation of the trip to school can be found as:

$\sigma =\sqrt{2^{2}+4^{2}}=\sqrt{4+16}=\sqrt{20}=4.47$

Therefore, standard deviation of the entire trip is 4.47 minutes.

We need to find the probability such that $P(x>30)=P(z>0.447)=0.67$

Therefore, the required probability is 0.67.

(d) If average time to walk to school is 10 minutes, then overall average time for the trip will be 10+20 = 30 minutes.

(e) Standard deviation won't change it will remain 4.47

(f) The new probability will be:

$z=\frac{x-\mu }{\sigma }=\frac{30-90}{4.47}=0$

$P(x>30)=P(z>0)=0.5$

Therefore, probability will be 0.50.

6 0

3 years ago

Read 2 more answers

goran spent $38 on fruit at the grocey store.He spent a total of 40$ at the store.what percentage of the total he spend of the f

lakkis [162]

The answer should be 95% I hope this helps :)

5 0

3 years ago

Jacob Lee is a frequent traveler between Los Angeles and San Diego. For the past month, he wrote down the flight times in minute

valkas [14]

Solution :

We know that

$H_0: \mu_1 = \mu_2=\mu_3$

$H_1 :$ At least one mean is different form the others (claim)

We need to find the critical values.

We know k = 3 , N = 35, α = 0.05

d.f.N = k - 1

= 3 - 1 = 2

d.f.D = N - k

= 35 - 3 = 32

SO the critical value is 3.295

The mean and the variance of each sample :

Goust Jet red Cloudtran

$\overline X_1 =50.5$ $\overline X_2 =50.07143$ $\overline X_3 =55.71429$

$s_1^2=19.96154$ $s_2^2=14.68681$ $s_3^2=36.57143$

The grand mean or the overall mean is(GM) :

$\overline X_{GM}=\frac{\sum \overline X}{N}$

$=\frac{51+51+...+49+49}{35}$

= 52.1714

The variance between the groups

$s_B^2=\frac{\sum n_i\left( \overline X_i - \overline X_{GM}\right)^2}{k-1}$

$=\frac{\left[14(50.5-52.1714)^2+14(52.07143-52.1714)^2+7(55.71426-52.1714)^2\right]}{3-1}$

$=\frac{127.1143}{2}$

= 63.55714

The Variance within the groups

$s_W^2=\frac{\sum(n_i-1)s_i^2}{\sum(n_i-1)}$

$=\frac{(14-1)19.96154+(14-1)14.68681+(7-1)36.57143}{(14-1)+(14-1)+(7-1)}$

$=\frac{669.8571}{32}$

= 20.93304

The F-test statistics value is :

$F=\frac{s_B^2}{s_W^2}$

$=\frac{63.55714}{20.93304}$

= 3.036212

Now since the 3.036 < 3.295, we do not reject the null hypothesis.

So there is no sufficient evidence to support the claim that there is a difference among the means.

The ANOVA table is :

Source Sum of squares d.f Mean square F

Between 127.1143 2 63.55714 3.036212

Within 669.8571 32 20.93304

Total 796.9714 34

3 0

3 years ago

PLEASE Answer! I'll Give Brainliest And A Thanks For A GOOD Answer.

melisa1 [442]

Answer: Your welcome the answer is in the attached Image.

Step-by-step explanation: mark me brainliest.

5 0

3 years ago