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3 years ago

Answer this pls

Mathematics

1 answer:

Charra [1.4K]3 years ago

6 0

Answer:

9g÷4x

Step-by-step explanation:

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Dan is watching The Birds in his backyard. Of the birds he watches, 9. Of them,or 45%, are sparrows. How many birds are in his b

ra1l [238]

If 9 of the sparrows is 45% of the birds in the backyard, 9/x = 45/100.

Cross multiply to get 9 * 100 = 45x; simplified is 900 = 45x or 20.

This means that there are 20 birds in the backyard. You can check this by dividing 9/20 which equals 45%

5 0

3 years ago

PLEASE HELP I WILL GIVE BRAINLIEST<br> Find the value of x in this case-

GenaCL600 [577]

Already answered here: brainly.com/question/26946183

4 0

2 years ago

Graph the solution set to this inequality.<br> 3x - 11 > 7x + 9

Bess [88]

Answer:

x<-5 <--------o open circle

Step-by-step explanation:

8 0

3 years ago

While sailing, Jacob is 150 feet from a lighthouse. The angle of elevation from his feet on the boat (at sea level) to the top o

Vlada [557]

Answer:

The height of the lighthouse is approximately 166.6 feet.

Step-by-step explanation:

Let the height of the lighthouse be represented by s, then;

Tan 48° = (opposite) ÷ (adjacent)

Tan 48° = s ÷ 150

⇒ s = 150 × Tan 48°

= 150 × 1.1106

= 166.59

s ≅ 166.6 feet

Therefore, the height of the lighthouse is approximately 166.6 feet.

7 0

3 years ago

In an investigation of pregnancy-induced hypertension, one group of women with this disorder was treated with low-dose aspirin,

AleksandrR [38]

Answer:

Step-by-step explanation:

Given that in an investigation of pregnancy-induced hypertension, one group of women with this disorder was treated with low-dose aspirin, and a second group was given a placebo. A sample consisting of 50 women who received aspirin has mean arterial blood pressure 120mmHg and standard deviation 10mmHg; a sample of 42 women who were given the placebo has mean blood pressure 115mmHg and standard deviation 12mmHg.

Population variances are equal

$H_0: \bar x = \bar y\\H_a: \bar x \neq \bar y$

(two tailed test at 5% significance level)

Here x denotes group I and Y group II

N Mean StDev SE Mean

Sample 1 50 120 10 1.4142

Sample 2 42 115 12 1.8516

Pooled std deviation = 10.9565

df=80

Mean difference= 5

Test statistic t = mean diff/std error = $\frac{5}{2.3299}$ =2.1803

p value = 0.0318

since p <0.05 we reject H0

b) We find p >0.01 hence at 1% significance level we accept H0

This implies that 99% confidence interval contains mean difference =0

4 0

4 years ago