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3 years ago

Which model shows two equal expressions when c= 4?

Mathematics

1 answer:

lapo4ka [179]3 years ago

3 0

Answer:

the third answer

Step-by-step explanation:

c = 4

c + 1 = 1 + 1 + 1 + 1 + 1

4 + 1 = 1 + 1 + 1 + 1 + 1

5 = 5

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What percent of the shaded checker board squares have pieces?

Alenkasestr [34]

Answer:

25%

Step-by-step explanation:

From the question given above, the following data were obtained:

Total shaded checker board (T) = 32

Shaded checker board with pieces (S) = 8

Percentage of Shaded checker board with pieces =?

We can obtain the percentage of checker board with pieces as follow:

Percentage of Shaded checker board with pieces = Shaded checker board with pieces / Total shaded checker board × 100

= S / T × 100

= 8 / 32 × 100

= 25%

Therefore, percentage of Shaded checker board with pieces is 25%

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

Stacy, Oliver, and Jivesh each plan to put a certain amount of money into their savings accounts that earn simple interest of 6%

Free_Kalibri [48]

Jivesh put $50
Stacy put $600
<span>Oliver put $100</span>

5 0

3 years ago

Read 2 more answers

5a +4b+c-3a-c+4b simplified

MAVERICK [17]

Answer:

2a + 8b

Step-by-step explanation:

4 0

3 years ago

A new company is in the process of evaluating its customer service. The company offers two types of sales: (1) Internet sales an

iragen [17]

Answer:

$H_A: P_{\text{Internet}}-P_{\text{Store}} > 0.10$

Step-by-step explanation:

We are given the following in the question:

Let $P_{\text{Internet}}$ be the proportion of the internet sales and $P_{\text{Store}}$ be the proportion of the store sale.

Hypothesis:

We have to conduct a hypothesis to check that the Internet sales are more than 10 percent higher than store sales.

Thus, we can design the null and alternative hypothesis as:

$H_{0}: P_{\text{Internet}}-P_{\text{Store}}\leq 0.10\\H_A: P_{\text{Internet}}-P_{\text{Store}} > 0.10$

Alternate Hypothesis:

The alternate hypothesis states that the proportion of the internet sales is greater than the proportion of store sales by 10 percent.

3 0

4 years ago