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2 years ago

Write the number shown in standard notation. 1.02 x 10 A. 1,020,000 B. 102,000,000 C. 102,000 D. 10,200,000

Mathematics

2 answers:

vitfil [10]2 years ago

8 0

Your answer should be A please mark brainlest

Irina18 [472]2 years ago

7 0

i think its c

Step-by-step explanation:

don't count me on this

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The life of a certain brand of battery is normally distributed, with mean 128 hours and standard deviation 16 hours. What is the

Katyanochek1 [597]

Answer:

x = 100 is equivalent to z = -1.75

Step-by-step explanation:

7 0

3 years ago

I don't get this problem. Lila decided to make kore baked goods for the bake sale. She used 1 / 8 lb less flour to make bread th

Anvisha [2.4K]

Start by taking the information we know, and write equations to represent the realationships:

1) "1/8 lbs less flour to make bread than to make cookies"
$b=c-\frac{1}{8}$

2) "1/4 lb more flour to make cookies than to make brownies:
$c=w+\frac{1}{4}$

3) "she used 1/2 lb of flour ro make bread"
$b = \frac{1}{2}$

From here you can solve for $w$ using back substitution.

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3 years ago

Which expression is equivalent to 4x + 6y - 8x?

zhannawk [14.2K]

Step-by-step explanation:

The expression is 6y - 4x

1) 6y - 4x

2) 8y - 4x

3) 4x + 6y

4) 6x - 4y

The correct answer is option 1

7 0

3 years ago

Elijah brought a total of 16 pounds of peanuts and cashew nuts and paid $49.50 if peanuts cost $2.75 per pound and cashews nuts

maria [59]

Answer:

The number of pounds of cashew is 11

Step-by-step explanation:

Let P represent the peanut

Let C represent the cashew nut.

From question, we were told that Elijah brought a total of 16 pounds of peanuts and cashew nuts. This can be written as:

P + C = 16 (1)

The total cost = $49.50

Cost per peanut = $2.75

Cost per cashew = $3.25

The above can be represented as:

49.50 = 2.75P + 3.25C. (2)

From equation 1,

P + C = 16

P = 16 — C

Substitute the value of P into equation 2:

49.50 = 2.75P + 3.25C.

49.50 = 2.75(16 — C) + 3.25C

49.50 = 44 — 2.75C + 3.25C

49.50 = 44 + 0.5C

Collect like terms

49.50 — 44 = 0.5C

5.5 = 0.5C

Divide both side by 0.5

C = 5.5/0.5

C = 11

Therefore, the number of pounds of cashew is 11

8 0

3 years ago

An automobile manufacturer who wishes to advertise that one of its models achieves 30 mpg (miles per gallon) decides to carry ou

Katyanochek1 [597]

Answer:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

$p_v =P(t_{(5)}$

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

Step-by-step explanation:

Data given and notation

The mean and sample deviation can be calculated from the following formulas:

$\bar X =\frac{\sum_{i=1}^n x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i -\bar X)}{n-1}}$

$\bar X=29.35$ represent the sample mean

$s=1.365$ represent the sample standard deviation

$n=6$ sample size

$\mu_o =30$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the true mean is lower than 30 or no, the system of hypothesis are :

Null hypothesis: $\mu \geq 30$

Alternative hypothesis: $\mu < 30$

Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$ (1)

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{29.35-30}{\frac{1.365}{\sqrt{6}}}=-1.17$

P-value

We need to calculate the degrees of freedom first given by:

$df=n-1=6-1=5$

Since is a one-side left tailed test the p value would given by:

$p_v =P(t_{(5)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

We can say that at 5% of significance the true mean is not significantly less than 30 so then the claim that true average fuel efficiency is (at least) 30 mpg makes sense

5 0

3 years ago