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3 years ago

How do you turn 26.2 into a fraction

Mathematics

2 answers:

vodomira [7]3 years ago

6 0

26 1/5. Rewrite the decimal number as a fraction with 1 in the denominator

26.2=26.21

Multiply to remove 1 decimal places. Here, you multiply top and bottom by 101 = 10

26.2/1×10/10=262/10

Find the Greatest Common Factor (GCF) of 262 and 10, if it exists, and reduce the fraction by dividing both numerator and denominator by GCF = 2,

262÷2/10÷2=131/5

Simplify the improper fraction,

=26 1/5

cupoosta [38]3 years ago

3 0

26.2 as a fraction equals 262/10

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Sales Daria sells televisions. She earns a fixed amount for each television and an additional $30 if the buyer gets an extended

alexira [117]

Answer:

$70

Step-by-step explanation:

Daria's total income = fixed income + variable income

Let

Daria's Fixed income = x

additional $30 if the buyer gets an extended warranty

Daria's variable income = 30s

Where,

s = number of televisions with extended warranties she sells

If Daria sells 15 televisions with extended warranties, she earns \$1,500. Find her fixed income

Daria's total income = fixed income + variable income

1500 = x + 30(15)

1500 = x + 450

1500 - 450 = x

1,050 = x

x = $1,050

Daria's total fixed income = $1,050

How much is the fixed amount Daria for each television?

Fixed income per television = Total fixed income / number of television sold

= 1050 / 15

= 70

Fixed income per television = $70

7 0

3 years ago

See the screenshot below

11111nata11111 [884]

2×2+3÷3=7÷3=2.333333=2.3

6 0

3 years ago

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

Testing the hypothesis, it is found that:

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

The critical value is: $t_c = 1.74$

The decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$ .

Hence, the decision rule is:

If t < 1.74, we do not reject the null hypothesis.
If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

The parameters are:

$\overline{x}$ is the sample mean.
$\mu$ is the value tested at the null hypothesis.
s is the standard deviation of the sample.
n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

$t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864

3 0

2 years ago

Help me please ASAP!

Bumek [7]

$216 = 6^{4x+ 11}$

$6^3 = 6^{4x+ 11}$

$3 = 4x + 11$

$4x = 3 - 11$

$4x = -8$

$x = -2$

Answer: (c) x = -2

8 0

3 years ago

59. What happens to the graph of y = |x| when the equation changes to y = |x + 4? A The graph shifts up 4 units. The graph shift

marishachu [46]

Answer: Graph shifts 4 units to the left

Explanation:

I'm assuming you meant to say y = |x+4|

If so, then the graph shifts 4 units to the left. Replacing x with x+4 moves the xy axis 4 units to the right if we held the V shape in place (since each x is now 4 units larger). This gives the illusion the V shape is moving 4 units to the left.

Or you could look at the vertex point to see how it moves. On y = |x|, the vertex is at (0,0). It then moves to (-4,0) when we go to y = |x+4|

8 0

3 years ago