Two times the difference of a number and four is the same as five less than the number

Everett and Marie are going to make brownies for their family reunion. They want to make 4 times the amount the recipe makes. If

denis-greek [22]

8/3 cups or 2 2/3
2/3 × 4/1 = 8/3
if u divide 3 into 8 it gives u 2 times with 2 places left
the 2 places equal 2/3
2 whole OR 6/3 + 2/3 = 8/3 OR 2 and 2/3

3 0

3 years ago

A lock has four dials. On each dial are the digits 0 to 9 how many possible combinations are there

madam [21]

Answer:

10,000

Step-by-step explanation:

The combination consists of four digits, hence 0000 to 9999. There at 10^4 or 10,000 possible combinations (but only one that works).

7 0

2 years ago

What is the answer to this problem

Svetlanka [38]

Answer: P = $47,120.53

4 0

2 years ago

Equation in slope-intercept form:

Jobisdone [24]

Answer:

y = -x + 3

Step-by-step explanation:

Find the slope using the formula [ y2-y1/x2-x1 ]. We can use the points (0, 3) and (3, 0) to solve.

0-3/3-0

-3/3

-1

From the graph, the y-intercept is (0, 3). Input all the data we know into the slope intercept form expression [ y = mx + b ].

y = -x + 3

Best of Luck!

5 0

3 years ago

While conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modem

Kitty [74]

Answer:

We conclude that this is an unusually high number of faulty modems.

Step-by-step explanation:

We are given that while conducting a test of modems being manufactured, it is found that 10 modems were faulty out of a random sample of 367 modems.

The probability of obtaining this many bad modems (or more), under the assumptions of typical manufacturing flaws would be 0.013.

Let p = population proportion.

So, Null Hypothesis, $H_0$ : p = 0.013 {means that this is an unusually 0.013 proportion of faulty modems}

Alternate Hypothesis, $H_A$ : p > 0.013 {means that this is an unusually high number of faulty modems}

The test statistics that would be used here One-sample z-test for proportions;

T.S. = $\frac{\hat p-p}{\sqrt{\frac{p(1-p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = sample proportion faulty modems= $\frac{10}{367}$ = 0.027

n = sample of modems = 367

So, the test statistics = $\frac{0.027-0.013}{\sqrt{\frac{0.013(1-0.013)}{367} } }$

= 2.367

The value of z-test statistics is 2.367.

Since, we are not given with the level of significance so we assume it to be 5%. Now at 5% level of significance, the z table gives a critical value of 1.645 for the right-tailed test.

Since our test statistics is more than the critical value of z as 2.367 > 1.645, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that this is an unusually high number of faulty modems.

6 0

3 years ago